RD Sharma 2021 Solutions for Class 9 Maths Chapter 12 Congruent Triangles are provided here with simple step-by-step explanations. These solutions for Congruent Triangles are extremely popular among class 9 students for Maths Congruent Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2021 Book of class 9 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2021 Solutions. All RD Sharma 2021 Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 12.15:
Question 1:
In the given figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that segment DE || BC.
Answer:
It is given that
We have to prove that
Now considering the two triangles we have
In
(Given)
(Given)
We need to show to prove
.
Now
(Vertically opposite angle)
So by congruence criterion we have
So and
Then
, and
Hence from above conditions .
Page No 12.15:
Question 2:
In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, OR, and RP respectively. Prove that LN = MN.
Answer:
It is given that
And is the mid point of
So
And is the mid point of
So
And is the mid point of
So
We have to prove that
In we have
(Equilateral triangle)
Then
, and
, and
Similarly comparing and
we have
, and
And (Since N is the mid point of
)
So by congruence criterion, we have
Hence.
Page No 12.15:
Question 3:
Prove that the medians of an equilateral triangle are equal.
Answer:
We have to prove that the median of an equilateral triangle are equal.
Let be an equilateral triangle with
as its medians.
Let
In we have
(Since
similarly
)
(In equilateral triangle, each angle
)
And (common side)
So by congruence criterion we have
This implies that,
Similarly we have
Hence .
Page No 12.15:
Question 4:
In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Answer:
In, it is given that
, and
We have to find, and
Since and
Then (as AB = AC)
Now
(By property of triangle)
Thus,
, as
(given)
So,
Since,, so
Hence .
Page No 12.15:
Question 5:
In a ΔABC, if AB = AC and ∠B = 70°, find ∠A.
Answer:
In it is given that
, and
We have to find.
Since
Then (isosceles triangles)
Now
(As
given)
Thus
(Property of triangle)
Hence .
Page No 12.15:
Question 6:
The vertical angle of an isosceles triangle is 100°. Find its base angles.
Answer:
Suppose in the isosceles triangle ΔABC it is given that
We have to find the base angle.
Now vertical angle (given)
And
Since then
Now
(By property of triangle)
So
Hence the base angle is.
Page No 12.15:
Question 7:
In the given figure, AB = AC and ∠ACD = 105°, find ∠BAC.
Answer:
It is given that
We have to find.
(Isosceles triangle)
Now
Since exterior angle of isosceles triangle is the sum of two internal base angles
Now
So, (By property of triangle)
Hence .
Page No 12.15:
Question 8:
Find the measure of each exterior angle of an equilateral triangle.
Answer:
We have to find the measure of each exterior angle of an equilateral triangle.
It is given that the triangle is equilateral
So, and
Since triangle is equilateral
So,
Now we have to find the exterior angle.
As we know that exterior angle of the triangle is sum of two interior angles
Thus
Hence each exterior angle is.
Page No 12.15:
Question 9:
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Answer:
It is given that the base of an isosceles triangle is produced on both sides.
We have to prove that the exterior angles so formed are equal to each other.
That is we need to show that
Let the is isosceles having base
and equal sides AB and AC
Then, and
(Isosceles triangles)
Now
.........(1)
And,
.......(2)
Thus
........(3)
Now from equation (2)
.........(4)
Since
Hence from equation (3) and (4)
Page No 12.15:
Question 10:
In the given figure, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
Answer:
It is given that
We have to find the ratio.
Since
And
So we have,
So
Hence .
Page No 12.16:
Question 11:
Determine the measure of each of the equal angles of a right-angled isosceles triangle.
Answer:
It is given that
Is right angled triangle
And
We have to find and
Since
(Isosceles triangle)
Now
(Property of triangle)
(
)
So
Hence
Page No 12.16:
Question 12:
In the given figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°
Answer:
It is given that
Δis a square and Δ
is an equilateral triangle.
We have to prove that
(1) and (2)
(1)
Since,
(Angle of square)
(Angle of equilateral triangle)
Now, adding both
Similarly, we have
Thus in and
we have
(Side of square)
And (equilateral triangle side)
So by congruence criterion we have
Hence.
(2)
Since
QR = RS ( Sides of Square)
RS = RT (Sides of Equilateral triangle)
We get
QR = RT
Thus, we get
(Angles opposite to equal sides are equal)
Now, in the triangle TQR, we have
Page No 12.16:
Question 13:
In the given figure, AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.
Answer:
It is given that
P and Q are equidistant from A and B that is
, and
We are asked to show that line PO is perpendicular bisector of line AB.
First of all we will show that ΔAQP and ΔQBP are congruent to each other and ultimately we get the result.
Consider the triangles AQP and QBP in which
AP=BP, AQ=BQ, PQ=PQ
So by SSS property we have
Implies that
Now consider the triangles ΔAPC and ΔPCB in which
And
So by SAS criterion we find that,
So this implies that AC=BC and
But
Hence PQ is perpendicular bisector of AB.
Page No 12.16:
Question 14:
In a âABC, D is the mid-point of AC such that BD = AC. Show that ∠ABC is a right angle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In a âABC, D is the mid-point of AC such that BD = AC.
D is the mid-point of AC.
∴ AD = CD = AC
⇒ AD = CD = BD (BD = AC)
In âABD,
AD = BD
∴ ∠ABD = ∠A .....(1) (In a triangle, equal sides have equal angles opposite to them)
In âCBD,
CD = BD
∴ ∠CBD = ∠C .....(2) (In a triangle, equal sides have equal angles opposite to them)
Adding (1) and (2), we get
∠ABD + ∠CBD = ∠A + ∠C
⇒ ∠B = ∠A + ∠C .....(3)
In âABC,
∠A + ∠B + ∠C = 180º (Angle sum property of triangle)
⇒ ∠B + ∠B = 180º [Using (3)]
⇒ 2∠B = 180º
⇒ ∠B = = 90º
Thus, ∠ABC is a right angle.
Page No 12.16:
Question 15:
âABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC,
AB = AC (Given)
∴ ∠C = ∠B .....(1) (In a triangle, angles opposite to equal sides are equal)
Also, ∠A = 90º (Given)
Now,
∠A + ∠B + ∠C = 180º (Angle sum property of triangle)
⇒ 90º + 2∠B = 180º [Using (1)]
⇒ 2∠B = 180º − 90º = 90º
⇒ ∠B = = 45º
∴ ∠C = ∠B = 45º
It is given that, AD is the bisector of ∠A.
∴ ∠CAD = ∠BAD = 45º
In âABD,
∠B = ∠BAD (Each measure 45º)
∴ AD = BD .....(2) (In a triangle, sides opposite to equal angles are equal)
In âACD,
∠C = ∠CAD (Each measure 45º)
∴ AD = CD .....(3) (In a triangle, sides opposite to equal angles are equal)
Adding (2) and (3), we have
AD + AD = BD + CD
⇒ 2AD = BC
Or BC = 2AD
Hence proved.
Page No 12.16:
Question 16:
âABC is a right triangle right angled at B such that ∠BCA = 2∠BAC. Show that AC = 2BC.
Answer:
![](/img/site_content/ask-answer/loader.gif)
âABC is a right triangle right angled at B such that ∠BCA = 2∠BAC.
Produce CB to D such that BC = BD.
In âABD and âABC,
BD = BC (Construction)
∠ABD = ∠ABC (90º each)
AB = AB (Common)
∴âABD ≅ âABC (SAS congruence axiom)
So, AD = AC .....(1) (CPCT)
∠BAD = ∠BAC .....(2) (CPCT)
Now,
∠BCA = 2∠BAC
⇒ ∠BCA = ∠BAC + ∠BAC
⇒ ∠BCA = ∠BAC + ∠BAD [Using (2)]
⇒ ∠BCA = ∠CAD
In âACD,
∠DCA = ∠CAD (Proved above)
⇒ AD = CD (Sides opposite to equal angles in a triangle are equal)
⇒ AC = BC + BD [Using (1)]
⇒ AC = BC + BC (BC = BD)
⇒ AC = 2BC
Page No 12.25:
Question 1:
BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE.
Answer:
It is given that
Is bisector of
and
is bisector of
.
And is isosceles with
We have to prove that
If will be sufficient to prove to show that
Now in these two triangles
Since, so
Now as BD and CE are bisector of the respectively, so
, and
BC=BC
So by congruence criterion we have
Hence Proved.
Page No 12.25:
Question 2:
In the given figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ΔSAT.
Answer:
It is given that
We have to prove that
Now
In we have
(Isosceles triangle) .......(1)
Now we have
(Vertically opposite angles)
(Since
, given)
.......(2)
Subtracting equation (2) from equation (1) we have
Now in and
we have
(Given)
So all the criterion for the two triangles and
are satisfied to be congruent
Hence by congruence criterion we have
proved.
Page No 12.25:
Question 3:
Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.
Answer:
It is given that
We have to prove that the lines and
bisect at
.
If we prove that, then
We can prove and
bisects at
.
Now in and
(Given)
(Since
and
is transversal)
And (since
and
is transversal)
So by congruence criterion we have,
, so
Hence and
bisect each other at
.
Page No 12.47:
Question 1:
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Answer:
It is given that
We are asked to show that
Let us assume
, and
are right angled triangle.
Thus in and
, we have
And (given)
Hence by AAs congruence criterion we haveProved.
Page No 12.47:
Question 2:
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Answer:
We have to prove that is isosceles.
Let Δ be such that the bisector
of
is parallel to
The base, we have
(Corresponding angles)
(Alternate angle)
(Since
)
Hence is isosceles.
Page No 12.47:
Question 3:
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Answer:
In the triangle ABC it is given that the vertex angle is twice of base angle.
We have to calculate the angles of triangle.
Now, let be an isosceles triangle such that
Then
(Given)
(
)
Now (property of triangle)
Hence
Page No 12.47:
Question 4:
Prove that each angle of an equilateral triangle is 60°
Answer:
We have to prove each angle of an equilateral triangle is.
Here
(Side of equilateral triangle)
...........(1)
And
(Side of equilateral triangle)
..........(2)
From equation (1) and (2) we have
Hence
Now
That is (since
)
Hence Proved.
Page No 12.47:
Question 5:
Angles A, B, C of a triangle ABC are equal to each other. Prove that ΔABC is equilateral.
Answer:
It is given that
We have to prove that triangle ΔABC is equilateral.
Since (Given)
So, ..........(1)
And (given)
So ........(2)
From equation (1) and (2) we have
Now from above equation if we have
Given condition satisfy the criteria of equilateral triangle.
Hence the given triangle is equilateral.
Page No 12.47:
Question 6:
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Answer:
It is given that
We have to find and
.
Since so,
Now (property of triangle)
(Since
)
Here
Then
Hence
Page No 12.47:
Question 7:
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Answer:
It is given that
We have to prove
In we have
(Given)
So,
Now (Given)
Since corresponding angle are equal, so
That is,
Henceproved.
Page No 12.47:
Question 8:
In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced prove that ∠MOC = ∠ABC.
Answer:
It is given that
In,
We have to prove that
Now
(Given)
Thus
........(1)
In, we have
So, {from equation (1)}
Hence Proved.
Page No 12.47:
Question 9:
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Answer:
In the following figure it is given that sides AB and PQ are parallel and BP is bisector of
We have to prove that is an isosceles triangle.
(Since BP is the bisector of
) ........(1)
(Since
and
are parallel) .......(2)
Now from equation (1) and (2) we have
So
Now since and
is a side of
.
And since two sides and
are equal, so
Hence is an isosceles triangle.
Page No 12.47:
Question 10:
ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
Answer:
It is given that in
And bisects
We have to prove that
Now let
(Given)
Sinceis a bisector of
so let
Let be the bisector of
If we join we have
In
So
In triangle and
we have
(Given)
(Proved above)
So by congruence criterion, we have
And
, and
(since
)
In we have
Since,
And,
So,
In we have
Here,
HenceProved.
Page No 12.47:
Question 11:
Bisectors of angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.
Answer:
![](/img/site_content/ask-answer/loader.gif)
∠ABD is the external angle adjacent to ∠ABC.
âABC is an isosceles triangle.
AB = AC (Given)
∴ ∠C = ∠ABC .....(1) (In a triangle, equal sides have equal angles opposite to them)
Also, OB and OC are the bisectors of ∠B and ∠C, respectively.
Similarly,
In âBOC,
∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)
Now,
∠ABD + ∠ABC = 180º .....(5) (Linear pair)
From (4) and (5), we have
∠ABD + ∠ABC = ∠ABC + ∠BOC
⇒ ∠ABD = ∠BOC
Thus, the external angle adjacent to ∠ABC is equal to ∠BOC.
Page No 12.57:
Question 1:
In the given figure, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ΔCBA.
Answer:
It is given that
We have to prove that.
Now in triangles and
we have
(Given)
(Given)
So (common)
Each side of is equal to
.
Hence, by congruence criterion we have
Proved.
Page No 12.58:
Question 2:
In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Answer:
It is given that
and L, M, N are the mid points of sides
,
, and
respectively.
We have to prove that
Now using the mid point theorem, we have
And
Similarly we have
In triangle and
we have
(Proved above)
(Proved above)
And (common)
So, by congruence criterion, we have
And
Then
HenceProved.
Page No 12.61:
Question 1:
ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is-isosceles.
Answer:
We have to prove thatis isosceles.
Let and
be perpendicular from D on AB and AC respectively.
In order to prove that
We will prove that
Now in and
we have
(Since D is mid point of BC)
(Given)
So by congruence criterion we have
And
Hence is isosceles.
Page No 12.61:
Question 2:
ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ΔABC is isosceles.
Answer:
It is given that
, and
And.
We have to prove is isosceles.
To prove is isosceles we will prove
For this we have to prove
Now comparing and
we have
(Given)
(Common side)
So, by right hand side congruence criterion we have
So (since sides opposite to equal angle are equal)
Hence is isosceles.
Page No 12.61:
Question 3:
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
Answer:
Let P be a point within such that
We have to prove that P lies on the bisector of
In and
we have
(We have)
(Common)
So by right hand side congruence criterion, we have
So,
Hence P lies on the bisector of proved.
Page No 12.61:
Question 4:
In the given figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that
∠DAQ = ∠CBP.
Answer:
It is given that
, and
If and
We have to prove that
In triangles and
we have
(Since
given)
So
And (given)
So by right hand side congruence criterion we have
So
HenceProved.
Page No 12.62:
Question 5:
Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) The measure of each angle of an equilateral triangle is 60°.
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Page No 12.62:
Question 6:
Fill in the blanks in the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are .......
(ii) Angle opposite to equal sides of a triangle are .......
(iii) In an equilateral triangle all angles are ........
(iv) In a ABC if ∠A = ∠C, then AB = ............
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ........
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ...... CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ΔABC ≅ Δ .........
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Page No 12.62:
Question 7:
ABCD is a square, X and Y are points on sides AD and BC respectively such that AY= BX. Prove that BY = AX and ∠BAY = ∠ABX.
Answer:
It is given ABCD is a square and
We have to prove that and
In right angled trianglesand Δ
we have
And, and
So by right hand side congruence criterion we have
So (since triangle is congruent)
HenceProved.
Page No 12.62:
Question 8:
ABCD is a quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In quadrilateral ABCD, AB = AD and BC = CD. Let AC and BD intersect at O.
In âABC and âADC,
AB = AD (Given)
AC = AC (Common)
BC = CD (Given)
∴ âABC ≅ âADC (SSS congruence criterion)
⇒ ∠BAC = ∠DAC (CPCT)
Now, in âABO and âADO,
AB = AD (Given)
∠BAO = ∠DAO (Proved above)
AO = AO (Common)
∴ âABO ≅ âADO (SAS congruence axiom)
⇒ OB = OD .....(1) (CPCT)
∠AOB = ∠AOD .....(2) (CPCT)
Now,
∠AOB + ∠AOD = 180º (Linear pair)
⇒ 2∠AOB = 180º [Using (2)]
⇒ ∠AOB = 90º .....(3)
From (1) and (3), we conclude that AC is the perpendicular bisector of BD.
Page No 12.62:
Question 9:
âABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
Answer:
![](/img/site_content/ask-answer/loader.gif)
âABC is a right triangle such that AB = AC. CD is the bisector of ∠C which intersects AB at D.
Let AB = AC = x units
In right âABC,
(Pythagoras theorem)
In âABC, CD is the bisector of ∠C.
(The bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle)
Now,
Hence proved.
Page No 12.62:
Question 10:
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that âOCD is an isosceles triangle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
O is a point in the interior of square ABCD. âOAB is an equilateral triangle.
Now,
∠DAB = ∠CBA .....(1) (Measure of each angle of a square is 90º)
∠OAB = ∠OBA .....(2) (Measure of each angle of an equilateral triangle is 60º)
Subtracting (2) from (1), we get
∠DAB − ∠OAB = ∠CBA − ∠OBA
⇒ ∠OAD = ∠OBC
In âOAD and âOBC,
OA = OB (Sides of an equilateral triangle are equal)
∠OAD = ∠OBC (Proved above)
AD = BC (Sides of a square are equal)
∴ âOAD ≅ âOBC (SAS congruence axiom)
⇒ OD = OC (CPCT)
In âOCD,
OC = OD
∴ âOCD is an isosceles triangle. (A triangle whose two sides are equal is an isosceles triangle)
Page No 12.62:
Question 11:
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ∠ABC and ∠ADC.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In quadrilateral ABCD, AB = BC and AD = CD.
In âABD and âCBD,
AB = CB (Given)
BD = BD (Common)
AD = CD (Given)
∴ âABD ≅ âCBD (SSS congruence criterion)
So, ∠ABD = ∠CBD .....(1) (CPCT)
∠ADB = ∠CDB .....(2) (CPCT)
From (1) and (2), we conclude that
BD bisects both ∠ABC and ∠ADC.
Page No 12.62:
Question 12:
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
Answer:
![](/img/site_content/ask-answer/loader.gif)
ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.
Join AN and BN.
In âAMN and âBMN,
AM = BM (M is the mid-point of AB)
∠AMN = ∠BMN (MN ⊥ AB)
MN = MN (Common)
∴ âAMN ≅ âBMN (SAS congruence axiom)
So, AN = BN .....(1) (CPCT)
∠ANM = ∠BNM (CPCT)
Now,
∠DNM = ∠CNM (90º each)
∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM
⇒ ∠AND = ∠BNC .....(2)
In âAND and âBNC,
DN = CN (N is the mid-point of CD)
∠AND = ∠BNC [From (2)]
AN = BN [From (1)]
∴ âAND ≅ âBNC (SAS congruence axiom)
So, AD = BC (CPCT)
Hence proved.
Page No 12.81:
Question 1:
In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Answer:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And is shortest side
Hence Is longest and
is shortest.
Page No 12.81:
Question 2:
In a ΔABC, if ∠B = ∠C = 45°, which is the longest side?
Answer:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since
)
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Page No 12.81:
Question 3:
In Δ ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD (ii) AD>AC
Answer:
It is given that
, and
We have to prove that
(1)
(2)
(1)
Now
And since BD=BC, so, and
That is,
Now
And, so
Hence (1) (Side in front of greater angle will be longer)
And (2) Proved.
Page No 12.81:
Question 4:
Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Answer:
As we know that a triangle can only be formed if
The sum of two sides is greater than the third side.
Here we have 2 cm, 3 cm and 7 cm as sides.
If we add
(Since 5 is less than 7)
Hence the sum of two sides is less than the third sides
So, the triangle will not exist.
Page No 12.81:
Question 5:
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Answer:
It is given that
AP is the bisector of
We have to arrange,
and
in descending order.
In we have
(As AP is the bisector of
)
So (Sides in front or greater angle will be greater) ........(1)
In we have
(As AP is the bisector of
)
Since,
So ..........(2)
Hence
From (1) & (2) we have
Page No 12.81:
Question 6:
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Answer:
We have to prove that the perimeter of a triangle is greater than the sum of its altitude.
In
,
,
We have to prove
Since
So and
By adding, we have
........(1)
Now consider then
, and
Now by adding .......(2)
Again consider
, and
By adding ...........(3)
Adding (1), (2) and (3), we get
Hence the perimeter of a triangle is greater than the sum of all its altitude.
Page No 12.81:
Question 7:
In the given figure, prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Answer:
(1) We have to prove that
In we have
(As sum of two sides of triangle is greater than third one) ........(1)
In we have
(As sum of two sides of triangle is greater than third one) .........(2)
Hence
Adding (1) & (2) we get
Proved.
(2) We have to prove that
In we have
(As sum of two sides of triangle is greater than third one)
(Adding
both sides)
Proved.
Page No 12.81:
Question 8:
Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
Page No 12.82:
Question 9:
Fill in the blanks to make the following statements true.
(i) In a right triangle the hypotenuse is the ...... side.
(ii) The sum of three altitudes of a triangle is ....... than its perimeter.
(iii) The sum of any two sides of a triangle is ..... than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ..... side opposite to it.
(v) Difference of any two sides of a triangle is ...... than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ...... angle opposite to it.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
Page No 12.82:
Question 10:
O is any point in the interior of Δ ABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > (AB + BC + CA)
Answer:
It is given that, is any point in the interior of
We have to prove that
(1) Produced
to meet
at
.
In we have
.........(1)
And in we have
.........(2)
Adding (1) & (2) we get
HenceProved.
(2) We have to prove that
From the first result we have
..........(3)
And
.........(4)
Adding above (4) equation
HenceProved.
(3) We have to prove that
In triangles,
and
we have
Adding these three results
HenceProved.
Page No 12.82:
Question 11:
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
Answer:
We have to prove that the sum of four sides of quadrilateral is greater than sum of diagonal.
Since the sum of two sides of triangle is greater than third side.
In we have
..........(1)
In we have
..........(2)
In we have
.........(3)
In we have
.........(4)
Adding (1) & (2) & (3) and (4) we get
Hence Proved.
Page No 12.82:
Question 12:
Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than of a right angle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Let AC be the longest side in the âABC.
Now,
AC > AB
⇒ ∠B > ∠C .....(1) (In a triangle, greater side has greater angle opposite to it)
Also,
AC > BC
⇒ ∠B > ∠A .....(2) (In a triangle, greater side has greater angle opposite to it)
From (1) and (2), we have
∠B + ∠B > ∠A + ∠C
⇒ ∠B + ∠B + ∠B > ∠A + ∠B + ∠C
⇒ 3∠B > 180º (Using angle sum property of a triangle)
⇒ ∠B > × 180º
Or ∠B > × 90º
Thus, the angle opposite to the longest side is greater than of a right angle.
Hence proved.
Page No 12.82:
Question 13:
D is any point on side AC of a âABC with AB = AC. Show that CD < BD.
Answer:
It is given that, D is any point on side AC of a âABC with AB = AC.
![](/img/site_content/ask-answer/loader.gif)
In âABC,
AB = AC (Given)
∴ ∠ACB = ∠ABC (In a triangle, equal sides have equal angles opposite to them)
Now, ∠ABC > ∠DBC
⇒ ∠ACB > ∠DBC (∠ACB = ∠ABC)
In âBCD,
∠DCB > ∠DBC
⇒ BD > CD (In a triangle, greater angle has greater side opposite to it)
Or CD < BD
Page No 12.84:
Question 1:
Mark the correct alternative in each of the following:
If ABC ΔLKM, then side of ΔLKM equal to side AC of ΔABC is
(a) LK
(b) KM
(c) LM
(d) None of these
Answer:
It is given that
As triangles are congruent, same sides will be equal.
So
Hence (c).
Page No 12.84:
Question 2:
If ΔABC ΔABC is isosceles with
(a) AB = AC
(b) AB = BC
(c) AC = BC
(d) None of these
Answer:
It is given that and
is isosceles
Since triangles are congruent so as same side are equal
Hence (a).
Page No 12.84:
Question 3:
If ΔABC ΔPQR and ΔABC is not congruent to ΔRPQ, then which of the following is not true:
(a) BC = PQ
(b) AC = PR
(c) AB = PQ
(d) QR = BC
Answer:
If and
is not congruent to
Since and compare corresponding sides you will see
(As
)
Hence (a) , is not true.
Page No 12.84:
Question 4:
In triangles ABC and PQR three equality relations between some parts are as follows:
AB = QP, ∠B = ∠P and BC = PR
State which of the congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Answer:
In and
It is given that
Since two sides and an angle are equal so it obeys
Hence (a).
Page No 12.84:
Question 5:
In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Answer:
In and
It is given that
Since given two sides and an angle are equal so it obeys
Hence (b).
Page No 12.84:
Question 6:
In ΔPQR ΔEFD then ED =
(a) PQ
(b) QR
(c) PR
(d) None of these
Answer:
If
We have to find
Since, as in congruent triangles equal sides are decided on the basis of “how they are named”.
Hence (c).
Page No 12.85:
Question 7:
If ΔPQR ΔEFD, then ∠E =
(a) ∠P
(b) ∠Q
(c) ∠R
(d) None of these
Answer:
If
Then we have to find
From the given congruence, as equal angles or equal sides are decided by the location of the letters in naming the triangles.
Hence (a)
Page No 12.85:
Question 8:
In a ΔABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Answer:
In the triangle ABC it is given that
We have to find
Now (linear pair)
Since
So, (by isosceles triangle)
This implies that
Now,
(Property of triangle)
Hence (a).
Page No 12.85:
Question 9:
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
(a) 100°
(b) 120°
(c) 110°
(d) 130°
Answer:
Let be isosceles triangle
Then
Now it is given that vertex angle is 2 times the sum of base angles
(As
)
Now
(Property of triangle)
(Since
, and
)
Hence (b).
Page No 12.85:
Question 10:
Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) SSA
(c) ASA
(d) SSS
Answer:
(b) ,as it does not follow the congruence criteria.
Page No 12.85:
Question 11:
In the given figure, the measure of ∠B'A'C' is
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Answer:
We have to find
Since triangles are congruent
So
Now in
(By property of triangle)
Hence (b) .
Page No 12.85:
Question 12:
If ABC and DEF are two triangles such that ΔABC ΔFDE and AB = 5cm, ∠B = 40°
(a) DF = 5cm, ∠F = 60°
(b) DE = 5cm, ∠E = 60°
(c) DF = 5cm, ∠E = 60°
(d) DE = 5cm, ∠D = 40°
Answer:
It is given thatand
So and
Now, in triangle ABC,
Therefore,
Hence the correct option is (c).
Page No 12.85:
Question 13:
In the given figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ΔABD is congruent to
(a) ΔEFC
(b) ΔECF
(c) ΔCEF
(d) ΔFEC
Answer:
It is given that
And
(Given)
(Given)
So (from above)
Hence
From (d)
.
Page No 12.86:
Question 14:
In the given figure, if AE || DC and AB = AC, the value of ∠ABD is
(a) 70°
(b) 110°
(c) 120°
(d) 130°
Answer:
We have to find the value of in the following figure.
It is given that
(Vertically apposite angle)
Now (linear pair) …… (1)
Similarly (linear pair) …… (2)
From equation (1) we have
Now (same exterior angle)
(Interior angle)
Now
So
Since
Hence (b).
Page No 12.86:
Question 15:
In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
(a) 52°
(b) 76°
(c) 156°
(d) 104°
Answer:
We are given that;
, is isosceles
And
We are asked to find angle x
From the figure we have
Therefore,
Since, so
Now
Hence (d) .
Page No 12.86:
Question 16:
In the given figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Answer:
It is given that
, is bisector of
We are to find the side CD
Analyze the figure and conclude that
(As in the two triangles are congruent)
In
So
Hence (c) .
Page No 12.86:
Question 17:
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE
Answer:
It is given that,
and
are the mid points of the sides
,
and
respectively of
(By mid point theorem)
(As it is mid point)
Now in and
(Common)
(Mid point)
(Mid point)
Hence (d)
Page No 12.86:
Question 18:
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
(a) 55°
(b) 70°
(c) 35°
(d) 110°
Answer:
It is given that, AB=AC and Ad is the median of BC
We know that in isosceles triangle the median from he vertex to the unequal side divides it into two equal part at right angle.
Therefore,
(Property of triangle)
Hence (a) .
Page No 12.87:
Question 19:
In the given figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Answer:
In the following figure we are given
Where ABCD is a square and AXYZ is also a square
We are asked to find BY
From the above figure we have XY=YZ=AZ=AX
Now in the given figure
So,
Now inn triangle ΔAXB
So
Hence (c) .
Page No 12.87:
Question 20:
In the given figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is
(a) 72°
(b) 73°
(c) 74°
(d) 95°
Answer:
It is given that
,
AB = CD
We have to find
Now AB = CD
AB = BD
Now the triangle is isosceles
Let
So
Now
Since
Hence (a) .
Page No 12.87:
Question 1:
If AB = QR, BC = PR and CA = PQ, then triangle _________ ≅ triangle _________.
Answer:
It is given that, AB = QR, BC = PR and CA = PQ.
So, A ↔ Q, B ↔ R and C ↔ P
∴ âABC ≅ âQRP
If AB = QR, BC = PR and CA = PQ, then triangle __ABC__ ≅ triangle __QRP__.
Page No 12.87:
Question 2:
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __________.
Answer:
SAS congruence axiom states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.
In âABC, ∠A is included between the sides AB and AC.
In âDEF, ∠D is included between the sides DF and DE.
![](/img/site_content/ask-answer/loader.gif)
∴ âABC ≅ âDEF by SAS axiom if AC = DE.
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __AC = DE__.
Page No 12.87:
Question 3:
In âABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __________ but not __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
If two sides of a triangle are equal, then it is an isosceles triangle.
In âABC, AB = AC
∴ âABC is an isosceles triangle.
In âABC,
AB = AC (Given)
∴ ∠C = ∠B (In a triangle, equal sides have equal angles opposite to them)
It is given that, ∠C = ∠P and ∠B = ∠Q.
∴ ∠P = ∠Q
In âPQR,
∠P = ∠Q (Proved)
∴ QR = PR (In a triangle, equal sides have equal angles opposite to them)
So, âPQR is an isosceles triangle.
However, it cannot be proved that the corresponding sides of âABC are congruent to the corresponding sides of âPQR. Hence, the triangles are not congruent.
In âABC and âPQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __isosceles__ but not __congruent__.
Page No 12.87:
Question 4:
In âPQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âPQR,
∠P = ∠R (Given)
∴ QR = PQ (Sides opposite to the equal angles of a triangle are equal)
⇒ PQ = 4 cm (QR = 4 cm)
In âPQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ___4 cm___.
Page No 12.87:
Question 5:
In âABC, AB = AC and ∠B = 50°. Then, ∠C = __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC,
AB = AC (Given)
∴ ∠C = ∠B (Angles opposite to the equal sides of a triangle are equal)
⇒ ∠C = 50° (∠B = 50°)
In âABC, AB = AC and ∠B = 50°. Then, ∠C = __50°__.
Page No 12.87:
Question 6:
In âABC, AB = BC and ∠B = 80°. Then ∠A = __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC,
AB = BC (Given)
∴ ∠C = ∠A ..... (1) (Angles opposite to the equal sides of a triangle are equal)
Now,
∠A + ∠B + ∠C = 180° (Angle sum property of triangle)
∴ ∠A + 80° + ∠A = 180°
⇒ 2∠A = 180° − 80° = 100°
⇒ ∠A = = 50°
In âABC, AB = BC and ∠B = 80°. Then ∠A = __50°__.
Page No 12.87:
Question 7:
If in âPQR, ∠P = 70° and ∠R = 30°, then the longest side of âPQR is ___________.
Answer:
In âPQR,
(Angle sum property of a triangle)
![](/img/site_content/ask-answer/loader.gif)
So, ∠Q is the greatest angle in the âPQR.
We know that, in a triangle the greater angle has the longer side opposite to it.
∴ PR is the longest side of âPQR.
If in âPQR, ∠P = 70° and ∠R = 30°, then the longest side of âPQR is ___PR___.
Page No 12.87:
Question 8:
In âPQR, if ∠R > ∠Q, then __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âPQR,
∠R > ∠Q (Given)
∴ PQ > PR (In a triangle, the greater angle has the longer side opposite to it)
In âPQR, if ∠R > ∠Q, then ___PQ > PR___.
Page No 12.87:
Question 9:
D is a point on side BC of a âABC such that AD bisects ∠BAC. Then ________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC, AD is the bisector of ∠A.
∴ ∠BAD = ∠CAD .....(1)
We know that exterior angle of a triangle is greater than each of interior opposite angle.
In âABD,
∠ADC > ∠BAD
⇒ ∠ADC > ∠CAD [Using (1)]
In âADC,
∠ADC > ∠CAD
∴ AC > CD (In a triangle, the greater angle has the longer side opposite to it)
Similarly, AB > BD
D is a point on side BC of a âABC such that AD bisects ∠BAC. Then ___AC > CD and AB > BD___.
Page No 12.87:
Question 10:
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between _________ and _________.
Answer:
In a triangle, the sum of two sides is greater than the third side.
∴ 5 cm + 1.5 cm > Third side
Or Third side < 6.5 cm
In a triangle, the difference of two sides is less than the third side.
∴ 5 cm − 1.5 cm < Third side
Or Third side > 3.5 cm
So,
3.5 cm < Third side < 6.5 cm
Thus, the length of the third side of the triangle lies between 3.5 cm and 6.5 cm.
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between __3.5 cm__ and _6.5 cm _.
Page No 12.87:
Question 11:
If AD is a median of âABC, then the perimeter of âABC cannot be less than or equal to ___________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
AD is the median of the âABC.
In âABD,
AB + BD > AD .....(1) (In a triangle, the sum of any two sides is greater than the third side)
In âACD,
CD + CA > AD .....(2) (In a triangle, the sum of any two sides is greater than the third side)
Adding (1) and (2), we get
AB + BD + CD + CA > AD + AD
⇒ AB + BC + CA > 2AD (BC = BD + CD)
⇒ Perimeter of âABC > 2AD
Or the perimeter of âABC cannot be less than or equal to 2AD
If AD is a median of âABC, then the perimeter of âABC cannot be less than or equal to ___2AD___.
Page No 12.87:
Question 12:
If âPQR ≅ âEDF, then PR = ___________.
Answer:
If âPQR ≅ âEDF, then
P ↔ E, Q ↔ D and R ↔ F
So, PR = EF, PQ = ED and QR = DF (If two triangles are congruent, then their corresponding sides are congruent)
If âPQR ≅ âEDF, then PR = ____EF____.
Page No 12.87:
Question 13:
It is given that âABC ≅ âRPQ, then BC = __________.
Answer:
If âABC ≅ âRPQ, then
A ↔ R, B ↔ P and C ↔ Q
So, AB = RP, BC = PQ and CA = QR (If two triangles are congruent, then their corresponding sides are congruent)
It is given that âABC ≅ âRPQ, then BC = ___PQ___.
Page No 12.87:
Question 14:
In âABC and âPQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by __________ criterion.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC and âPQR,
∠A = ∠Q
∠B = ∠R
AC = PR
∴ âABC ≅ âQRP (AAS congruence criterion)
AAS congruence criterion states that if any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.
In âABC and âPQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by __AAS congruence__ criterion.
Page No 12.88:
Question 15:
In âABC and âPQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by _______ criterion.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âABC and âPQR,
∠A = ∠Q
∠B = ∠R
AB = QR
∴ âABC ≅ âQRP (ASA congruence criterion)
ASA congruence criterion states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
In âABC and âPQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by __ASA congruence criterion__.
Page No 12.88:
Question 1:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
Answer:
It is given that
Since, the triangles ABC and DEF are congruent, therefore,
Page No 12.88:
Question 2:
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C =∠F. Are the two triangles necessarily congruent?
Answer:
It is given that
For necessarily triangle to be congruent, sides should also be equal.
Page No 12.88:
Question 3:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5cm and ∠D = 75°. Are two triangles congruent?
Answer:
It is given that
Since, two sides and angle between them are equal, therefore triangle ABC and DEF are congruent.
Page No 12.88:
Question 4:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Answer:
The given information and corresponding figure is given below
From the figure, we have
And,
Hence, triangles ABC and ADC are congruent to each other.
Page No 12.88:
Question 5:
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°.
Are two triangles congruent?
Answer:
For the triangles ABC and ECD, we have the following information and corresponding figure:
In triangles ABC and ECD, we have
The SSA criteria for two triangles to be congruent are being followed. So both the triangles are congruent.
Page No 12.88:
Question 6:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Answer:
In the triangle ABC it is given that
,
and
are medians.
We have to show that
To show we will show that
In triangle ΔBFC and ΔBEC
As, so
.........(1)
BC=BC (common sides) ........(2)
Since,
As F and E are mid points of sides AB and AC respectively, so
BF = CE ..........(3)
From equation (1), (2), and (3)
HenceProved.
Page No 12.88:
Question 7:
Find the measure of each angle of an equilateral triangle.
Answer:
In equilateral triangle we know that each angle is equal
So
Now (by triangle property)
Hence.
Page No 12.88:
Question 8:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ ΔBCE.
Answer:
We have to prove that
Given is a square
So
Now in is equilateral triangle.
So
In and
(Side of triangle)
(Side of equilateral triangle)
And,
So
Hence from congruence
Proved.
Page No 12.89:
Question 9:
Prove that the sum of three altitudes of a triangle is less than the sum of its sides.
Answer:
We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.
In we have
,
and
We have to prove
As we know perpendicular line segment is shortest in length
Since
So ........(1)
And
........(2)
Adding (1) and (2) we get
........(3)
Now, so
.......(4)
And again, this implies that
........(5)
Adding (3) & (4) and (5) we have
HenceProved.
Page No 12.89:
Question 10:
In the given figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
Answer:
It is given that
, and
We have to prove that
We basically will prove to show
In and
(Given)
(Given)
And is common in both the triangles
So all the properties of congruence are satisfied
So
Hence Proved.
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