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#### Question 1:

In each of the following, write the correct answer:
Which of the following is not a criterion for congruence of triangles?
(A) SAS
(B) ASA
(C) SSA
(D) SSS

Two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.

Congruence of triangles are SAS (Side-Angle-Side), ASA (Angle-Side- Angle), SSS (Side-Side-Side) and RHS (right angle-hypotenuse-side).

SSA is not a criteria for congruency. If two triangles seems to be congruent by SSA rule, they cannot be said congruent.

#### Question 2:

In each of the following, write the correct answer:
If AB = QR, BC = PR and CA = PQ, then
(A) âˆ†ABC â‰Œ âˆ†PQR
(B) âˆ†CBA â‰Œ âˆ†PRQ
(C) âˆ†BAC â‰Œ âˆ†RPQ
(D) âˆ†PQR â‰Œ âˆ†BCA

Given that AB = QR, BC = PR and CA = PQ.
A↔Q, B↔R, C↔P

Under this correspondence,

$△\mathrm{ABC}\cong △\mathrm{QRP}$
or $△\mathrm{CBA}\cong △\mathrm{PRQ}$
Hence, the correct answer is option B.

#### Question 3:

In each of the following, write the correct answer:
In âˆ†ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(A) 40°
(B) 50°
(C) 80°
(D) 130°

Given: In $△\mathrm{ABC}$
AB = AC and $\angle \mathrm{B}$ = 50°

Since, AB = AC
$△\mathrm{ABC}$ is an Isosceles triangles.
$\therefore \angle \mathrm{B}=\angle \mathrm{C}$ ($\because$ Angle opposite to equal sides are equal)
$⇒\angle \mathrm{C}$ = 50°           $\left(\because \angle \mathrm{B}=50°\right)$

Hence, the correct answer is option B.

#### Question 4:

In each of the following, write the correct answer:
In âˆ†ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(A) 80°
(B) 40°
(C) 50°
(D) 100°

Given: In $△\mathrm{ABC}$
BC = AB and $\angle \mathrm{B}$ = 80°
Now,

$\angle \mathrm{A}=\angle \mathrm{C}$ (angle opposite to equal sides are equal)
Let $\angle \mathrm{A}=\mathrm{x}=\angle \mathrm{C}$
By angle sum property.
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$
$⇒x+80°+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=100°\phantom{\rule{0ex}{0ex}}⇒x=50°\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{A}=50°$
Hence, the correct answer is option C.

#### Question 5:

In each of the following, write the correct answer:
In âˆ†PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
(A) 4 cm
(B) 5 cm
(C) 2 cm
(D) 2.5 cm

In $△\mathrm{PQR}$

Since $\angle \mathrm{R}=\angle \mathrm{P}$
PQ = QR [Sides opposite to equal angles are equal].

Hence, the correct answer is option A.

#### Question 6:

In each of the following, write the correct answer:
D is a point on the side BC of a âˆ†ABC such that AD bisects ∠BAC. Then
(A) BD = CD
(B) BA > BD
(C) BD > BA
(D) CD > CA

Given: In  bisects $\angle \mathrm{BAC}$
$\therefore \angle \mathrm{BAD}=\angle \mathrm{DAC}$      .....(1)

Now, in  is a exterior angle.
$\therefore \angle \mathrm{BDA}>\angle \mathrm{DAC}$      [$\because$ Exterior angles is equal to sum of two interior angle]

Hence, the correct answer is option B.

#### Question 7:

In each of the following, write the correct answer:
It is given that âˆ†ABC â‰Œ âˆ†FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true?
(A) DF = 5 cm, ∠F = 60°
(B) DF = 5 cm, ∠E = 60°
(C) DE = 5 cm, ∠E = 60°
(D) DE = 5 cm, ∠D = 40°

Given:

$\therefore$ AB = DF  [by CPCT]
$⇒$DF = 5 cm
Also,
$\angle \mathrm{C}=\angle \mathrm{E}$ [by CPCT]
$⇒\angle \mathrm{C}=\angle \mathrm{E}=180-\left(\angle \mathrm{A}+\angle \mathrm{B}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{C}=\angle \mathrm{E}=180-\left(80°+40°\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{C}=\angle \mathrm{E}=60°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{E}=60°$
Hence, the correct answer is option B.

#### Question 8:

In each of the following, write the correct answer:
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(A) 3.6 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm

From triangle inequality property, the difference of two sides < third side and sum of two sides > third side.
∴ 5 − 1.5 < BC and 5 + 1.5 > BC
⇒ 3.5 < BC and 6.5 > BC
⇒ 3.5 < BC < 6.5

Hence, the correct answer is option D.

#### Question 9:

In each of the following, write the correct answer:
In âˆ†PQR, if ∠R > ∠Q, then
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

Given: $\angle \mathrm{R}>\angle \mathrm{Q}$

$\therefore \mathrm{PQ}>\mathrm{PR}$ (Side opposite to longer angle is larger)
Hence, the correct answer is option B.

#### Question 10:

In each of the following, write the correct answer:
In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(A) isosceles but not congruent
(B) isosceles and congruent
(C) congruent but not isosceles
(D) neither congruent nor isosceles

Given: In

$\therefore \angle \mathrm{A}=\angle \mathrm{R}$ (third angle of triangle)
Since, all three angles of both triangles are equal
$\therefore △\mathrm{ABC}~△\mathrm{PQR}$
Also, AB = AC (given)
$⇒\angle \mathrm{B}=\angle \mathrm{C}$

Thus, $△\mathrm{ABC}$ is an isosceles triangle.
But,

$⇒\angle \mathrm{P}=\angle \mathrm{Q}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PR}=\mathrm{QR}$

Thus, both triangles are isosceles but not congruent.
Hence, the correct answer is option A.

#### Question 11:

In each of the following, write the correct answer:
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if
(A) BC = EF
(B) AC = DE
(C) AC = EF
(D) BC = DE

Given: In
AB = FD, $\angle \mathrm{A}=\angle \mathrm{D}$

We know, two triangle will be congruent by SAS axiom, if two sides and the angle included between then are equal.
$\therefore$ AC = DE
Hence, the correct answer is option B.

#### Question 1:

In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of âˆ†PQR should be equal to side AB of âˆ†ABC so that the two triangles are congruent? Give reason for your answer.

Now, For $△\mathrm{ABC}\cong △\mathrm{QRP}$ by ASA axiom.
$⇒\mathrm{AB}=\mathrm{QR}$ (since, in ASA axiom, two angles and a side between them is included).

#### Question 2:

In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of âˆ†PQR should be equal to side BC of âˆ†ABC so that the two triangles are congruent? Give reason for your answer.

Now, for $△\mathrm{ABC}\cong △\mathrm{QRP}$ By AAS axiom
$⇒\mathrm{BC}=\mathrm{RP}$ (in AAS axiom, two angles and a side not between them is included)

#### Question 3:

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

False, because two triangles will be congruent when corresponding equal angles must be contained in the equal sides.

#### Question 4:

“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

True, because the triangle will be congruent either by ASA rule or AAS rule. This is because two angles and one side are sufficient to construct two congruent triangles.

#### Question 5:

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.

Given: Length of sides of triangle: 4 cm, 3 cm, 7 cm.
We know, sum of two sides of a triangle must be greater than third side.
But, if we add 4 + 3 = 7
Hence, it is not possible to construct a triangle with sides 4 cm, 3 cm and 7 cm.

#### Question 6:

It is given that âˆ†ABC â‰Œ âˆ†RPQ. Is it true to say that BC = QR? Why?

Given: $△\mathrm{ABC}\cong △\mathrm{RPQ}$

Since, both triangles are congruent,
$\therefore$ AB = PR, AC = RQ, BC = PQ
Hence, BC $\ne$QR.

#### Question 7:

If âˆ†PQR â‰Œ âˆ†EDF, then is it true to say that PR = EF? Give reason for your answer.

Given: $∆\mathrm{PQR}\cong △\mathrm{EDF}$

Since, both triangles are congruent,

Hence, it is true to say PR = EF.

#### Question 8:

In âˆ†PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is the longest? Give reason for your answer.

Given: In $△\mathrm{PQR}$,

By angle sum property, we have
$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180°\phantom{\rule{0ex}{0ex}}⇒70+\angle \mathrm{Q}+30°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{Q}=80°$
We know, side opposite to largest angle will be larger.
$\therefore$ PR will be the largest side in $△\mathrm{PQR}$.

#### Question 9:

AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2 AD? Give reason for your answer.

Given: AD is median of $△\mathrm{ABC}$

In $△\mathrm{ABD}$,
$\mathrm{AB}+\mathrm{BD}>\mathrm{AD}$        .....(1) [sum of any two sides of a triangle is greater than third side]
Similarly, in $△\mathrm{ACD}$
AC + CD > AD     .....(2)
Now, on adding (1) and (2), we get
AB + BD + AC + CD > 2AD
AB + BC + CA > 2AC          ($\therefore \mathrm{BD}+\mathrm{CD}=\mathrm{BC}$)

Hence, it is true.

#### Question 10:

M is a point on side BC of a triangle ABC such that AM is the bisector of ∠BAC. Is it true to say that perimeter of the triangle is greater than 2 AM? Give reason for your answer.

True,

In $△\mathrm{ABM}$
AB + BM > AM       .....(1) ($\therefore$ Sum of two sides of a triangle is greater than third side).
Similarly, in $△\mathrm{AMC}$

Adding (1) and (2), we get
AB + BM + CM + AC > 2AM
AB + BC + CA > 2AM       ($\therefore$ BM + CM = BC)

Hence, the perimeter of $△\mathrm{ABC}$ is greater than 2AM.

#### Question 11:

Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.

Given that the sides of triangle are 9 cm, 7 cm, 17 cm.
We know, the sum of any two sides of a triangle must be greater than the third side.
But 9 + 7 = 16 ()

Hence, it is not possible to construct a triangle.

#### Question 12:

Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.

Given that the sides of a triangle are 8 cm, 7 cm, 4 cm.
We know, sum of any two sides of a triangle must be greater than third sides.

Hence, it is possible to construct the triangle.

#### Question 1:

ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.

Given: $△\mathrm{ABC}$ is isosceles triangle
AB = AC, BD and CE are medians

Now, AB =  AC (given)
$⇒\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{AC}$
$⇒\mathrm{AE}=\mathrm{AD}$ ($\because$ E and D are midpoint of AB and AC)

In âˆ†ABD and âˆ†ACE,
AB = AC (given)
∠A = ∠A (common)
$\therefore ∆\mathrm{ABD}\cong ∆\mathrm{ACE}$ [by SAS]
$⇒$BD = CE (by CPCT)

Hence, proved.

#### Question 2:

In the Figure, D and E are points on side BC of a âˆ†ABC such that BD = CE and AD = AE. Show that âˆ†ABD â‰Œ âˆ†ACE.

Given: In $△\mathrm{ABC}$, BD = EC and AD = AE

To show: $△\mathrm{ABD}\cong △\mathrm{ACE}$
$\angle \mathrm{ADE}=\angle \mathrm{AED}$ (angle opposite to equal sides are equal)
Now $\angle \mathrm{ADB}+\angle \mathrm{ADE}=180°$

In

Hence, $△\mathrm{ABD}\cong △\mathrm{ACE}$ (by SAS)

#### Question 3:

CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that âˆ†ADE â‰Œ âˆ†BCE.

Given: CDE is an equilateral triangle formed on side CD of square ABCD.
To show: $△\mathrm{ADE}\cong △\mathrm{BCE}$
Proof:

Since, $△\mathrm{DEC}$ is equilateral,
$⇒$DC = EC = DE          .....(1)
and , $\angle \mathrm{CDE}=\angle \mathrm{DEC}=\angle \mathrm{DCE}=60°$    .....(2)
Also, ABCD is a square,
$⇒$AD = DC = CB = AB     .....(3)
and $\angle \mathrm{ABC}=\angle \mathrm{BCD}=\angle \mathrm{CDA}=\angle \mathrm{DAB}=90°$     .....(4)
Now,
$\begin{array}{rcl}\angle \mathrm{ADE}& =& \angle \mathrm{ADC}+\angle \mathrm{CDE}\\ & =& 90°+60°\\ & =& 150°\end{array}$
Similarly,
$\begin{array}{rcl}\angle \mathrm{BCE}& =& \angle \mathrm{BCD}+\angle \mathrm{DCE}\\ & =& 90°+60°\\ & =& 150°\end{array}$
$\therefore \angle \mathrm{ADE}=\angle \mathrm{BCE}$   .....(5)
In
(from 5)
DE = CE (given)
$\therefore △\mathrm{ADE}\cong △\mathrm{BCE}$ (by SAS)

#### Question 4:

In the Figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that âˆ†ABC â‰Œ âˆ†DEF.

Given: BA$\perp$AC, DE$\perp$DF such that BA = DE, BF = EC.
To show: $△\mathrm{ABC}\cong △\mathrm{DEF}$
Proof:

Here, BF = CE (given)
Adding FC on both sides we get
BF + FC = CE + FC
$⇒$BC = FE ............................(1).
Now In
$\angle \mathrm{A}=\angle \mathrm{D}$ (given)
AB = DE (given)
BC = EF (from 1)
$\therefore △\mathrm{ABC}\cong △\mathrm{DEF}$ (by RHS).

#### Question 5:

Q is a point on the side SR of a âˆ†PSR such that PQ = PR. Prove that PS > PQ.

Given: Q is a point on side SR of âˆ†PSR such that PQ = PR.
To prove: PS > PQ.
Proof:

Here, PQ = PR (given)
$\angle \mathrm{PQR}=\angle \mathrm{PRQ}$    .....(1) (angle opposite to equal sides are equal)
Now, in  is an exterior angle
$\therefore \angle \mathrm{PQR}>\angle \mathrm{PSQ}$ (exterior angle of a triangle is greater than each of opposite interior angle)
From equation (1) and (2)
$⇒\angle \mathrm{R}>\angle \mathrm{S}$
$⇒\mathrm{PS}>\mathrm{PR}$ [side opposite to greater angle is longer]

Hence proved.

#### Question 6:

S is any point on side QR of a âˆ†PQR. Show that: PQ + QR + RP > 2PS.

Given: In  is any point on QR
To show: PQ + QR + RP > 2PS
Proof:

In $△\mathrm{PQS}$,
PQ + QS > PS  .....(1) (sum of any two sides if greater than third side)
Also, in $△\mathrm{PRS}$
$\mathrm{PR}+\mathrm{RS}>\mathrm{PS}$  .....(2)
Adding (1) and (2), we get
PQ + (QS + RS) + PR > 2PS
⇒ PQ + QR + PR > 2PS    [$\because$QR = QS + SR]

Hence proved.

#### Question 7:

D is any point on side AC of a âˆ†ABC with AB = AC. Show that CD < BD.

Given: In $△\mathrm{ABC},$ D is any point on AC such that AB = AC.
To show: CD < BD
Proof:

In $△\mathrm{ABC}$
AB = AC (given)
$\angle \mathrm{ABC}=\angle \mathrm{ACB}$ (angle opposite to equal sides are equal)
In  since $\angle \mathrm{DBC}$ is an internal angle of $\angle \mathrm{ABC}$

Hence verified.

#### Question 8:

In the Figure, l || m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively.

Given: $l\parallel m$, M is midpoint of AB
To show: M is midpoint of CD
Proof:

In
$\angle \mathrm{AMC}=\angle \mathrm{BMD}$ [vertically opposite angles]
AM = MB (given)
$\angle \mathrm{CAM}=\angle \mathrm{MBD}$ (alternate interior angles)
$△\mathrm{AMC}\cong △\mathrm{DMB}$ [By ASA axiom]
$\therefore$ DM = MC (by CPCT)
Here, M is the mid-point of DC.

#### Question 9:

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that ∠MOC = ∠ABC.

Given: $△\mathrm{ABC}$ is isosceles with AB = AC
To prove: $\angle \mathrm{MOC}=\angle \mathrm{ABC}$
Proof:

In $△\mathrm{ABC}$,
AB = AC (given)
$\angle \mathrm{ABC}=\angle \mathrm{ACB}$ (angle opposite to equal sides are equal)
$⇒\frac{1}{2}\angle \mathrm{ABC}=\frac{1}{2}\angle \mathrm{ACB}$ (dividing both sides by 2)
$⇒\angle \mathrm{OBC}=\angle \mathrm{OCB}$    .....(1) ($\because$ OB and OC are bisectors of ).
Now,
(from (1))
$⇒\angle \mathrm{MOC}=\angle \mathrm{ABC}$
Hence, proved.

#### Question 10:

Bisectors of the 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.

Given: In  meet at O.
To prove: $\angle \mathrm{DBA}=\angle \mathrm{BOC}$
Proof:

In $△\mathrm{ABC}$,
AB = AC
$\angle \mathrm{ABC}=\angle \mathrm{ACB}$ (angle opposite to equal sides are equal).
$⇒\frac{1}{2}\angle \mathrm{ABC}=\frac{1}{2}\mathrm{ACB}$
$⇒\angle \mathrm{OBC}=\angle \mathrm{OCB}$   .....(1)
In $△\mathrm{BOC}$,
$\angle \mathrm{OBC}+\angle \mathrm{OCB}+\angle \mathrm{BOC}=180°$
$⇒2\angle \mathrm{OBC}+\angle \mathrm{BOC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ABC}+\angle \mathrm{BOC}=180°$
$⇒180°-\angle \mathrm{DBA}+\angle \mathrm{BOC}=180°$ [$\because$DBC is a straight line]
$⇒\angle \mathrm{BOC}=\angle \mathrm{DBA}$
Hence, proved.

#### Question 11:

In the Figure, AD is the bisector of ∠BAC. Prove that AB > BD.

Given: In $△\mathrm{ABC}$, AD is bisector $\angle \mathrm{BAC}$
To prove: AB > BD
Proof:

Here, $\angle \mathrm{BAD}=\angle \mathrm{DAC}$ ($\because$ AD is bisector of $\angle \mathrm{BAC}$)
Now, In $△\mathrm{ADC}$
$\angle \mathrm{ADB}=\angle \mathrm{DAC}+\angle \mathrm{C}$ (from exterior angle property)
$⇒\angle \mathrm{ADB}>\angle \mathrm{DAC}$ (Exterior angle is greater than each of interior angles)

$⇒\mathrm{AB}>\mathrm{BD}$ (side opposite to greater angle is greater)
Hence, proved.

#### Question 1:

Find all the angles of an equilateral triangle.

Given: $△\mathrm{ABC}$ is an equilateral triangle

Since all angles in an equilateral triangle are equal.
Let each angle be $x$.
$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=x$

$x+x+x=180°$
$⇒3x=180°\phantom{\rule{0ex}{0ex}}⇒x=60°$
Thus, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60°$.

#### Question 2:

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

To prove: AO = BO

We know,
Angle of incidence = Angle of reflection
$\angle \mathrm{ACN}=\angle \mathrm{DCN}$        .....(1)
Since, AB $\parallel$CN, and AC is transversal.
$\angle \mathrm{OAC}=\angle \mathrm{ACN}$        .....(2)
Also AB$\parallel$CN, BD as transversal.
$\angle \mathrm{DCN}=\angle \mathrm{ABC}$        .....(3)  (corresponding angles)
From (1), (2) and (3), we get

$\angle \mathrm{CBO}=\angle \mathrm{OAC}$      .....(4)

Now, in
$\angle \mathrm{BOC}=\angle \mathrm{AOC}$ [each 90°]
$\angle \mathrm{CBD}=\angle \mathrm{OAC}$ [from (4)]
CO = CO [common]
$\therefore △\mathrm{ACO}\cong △\mathrm{BCO}$     (from AAS axiom)
AO = BO (by CPCT)
Hence, proved.

#### Question 3:

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC . To prove that ∠BAD = ∠CAD, a student proceeded as follows:

In âˆ†ABD and âˆ†ACD,

AB = AC             (Given)
∠B = ∠C             (because AB = AC)
Therefore, âˆ†ABD â‰Œ âˆ†ACD     (AAS)
What is the defect in the above arguments?

In
AB = AC (given)
$\angle \mathrm{ADB}=\angle \mathrm{ADC}$ (each 90°)
$\therefore △\mathrm{ADB}\cong △\mathrm{ADC}$ (by RHS axiom)
$\angle \mathrm{BAD}=\angle \mathrm{CAD}$ (by CPCT)

#### Question 4:

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.

P is a point on bisector of $\angle \mathrm{ABC}$
To prove: $△\mathrm{BPQ}$ is an isosceles triangle.
Proof:

Here $\angle \mathrm{ABP}=\angle \mathrm{CBP}$   .....(1) (BP is angle bisector)
Now AB$\parallel$PQ, BP is a transversal
$\angle \mathrm{BPQ}=\angle \mathrm{ABP}$           .....(2)  (alternate interior angle)
From (1) and (2), we get
$\angle \mathrm{CBP}=\angle \mathrm{BPQ}$
BQ = PQ (side opposite to equal angles are equal)
Thus,  is an isosceles triangle.
Hence, proved.

#### Question 5:

ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

To show: BD bisects .
Proof:

In
AB = BC (given)
BD = BD (common)

Hence, BD bisects .

#### Question 6:

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

Given:  is right triangle, AB = AC
Proof:

In
AB = AC (given)
$\angle \mathrm{BAD}=\angle \mathrm{CAD}$ (given)
$△\mathrm{ABD}\cong △\mathrm{ACD}$ (by SAS axiom)
$\therefore \angle \mathrm{ADB}=\angle \mathrm{ADC}$ (by CPCT)

Also, $\angle \mathrm{ADB}=\angle \mathrm{ADC}=90°$
BD = DC (by CPCT)
In $△\mathrm{ABD}$,
${\left(\mathrm{AB}\right)}^{2}={\left(\mathrm{AD}\right)}^{2}+{\left(\mathrm{BD}\right)}^{2}$   .....(1)
In $△\mathrm{ADC}$,
${\left(\mathrm{AC}\right)}^{2}={\left(\mathrm{AD}\right)}^{2}+{\left(\mathrm{DC}\right)}^{2}$   .....(2)
Adding (1) and (2), we get
${\left(\mathrm{AB}\right)}^{2}+{\left(\mathrm{AC}\right)}^{2}=2{\left(\mathrm{AD}\right)}^{2}+{\left(\mathrm{BD}\right)}^{2}+{\left(\mathrm{BC}\right)}^{2}$

Hence, proved.

#### Question 7:

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that âˆ†OCD is an isosceles triangle.

Given: OAB is an equilateral triangle
To prove: $△\mathrm{OCD}$ is isosceles triangle.
Proof:

Here, $△\mathrm{OAB}$ is an equilateral triangle.
$\angle \mathrm{OAB}=\angle \mathrm{OBA}=60°$    .....(1)
$\angle \mathrm{DAB}=\angle \mathrm{CBA}=90°$    .....(2)
Subtracting (1) from (2), we get

Now, in
AO = BO (given)
(from (3))
AD = BC (ABCD is a square)
$△\mathrm{AOD}\cong △\mathrm{BOC}$ (by SAS)
$⇒\mathrm{CO}=\mathrm{OD}$ (by CPCT)
Thus, $△\mathrm{COD}$ is an isosceles triangle.

#### Question 8:

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Given:  on same base BC
AB = AC and DB = DC
To show: AD is perpendicular bisector of BC
Proof:

In
AB = AC (given)
BD = DC (given)

Now, in ,
AB = AC (given)
AO = AO (common)
$\angle \mathrm{BAO}=\angle \mathrm{CAO}$ (from (1))
$\therefore △\mathrm{AOB}\cong △\mathrm{AOC}$ (by SAS congruence rule)
BO = OC (by CPCT)
(by CPCT)
Now,$\angle \mathrm{AOB}+\angle \mathrm{AOC}=180°$ (by linear pair)
$⇒\angle \mathrm{AOB}=90°$
Hence, AD$\perp$BC and AD bisect BC.

#### Question 9:

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

Given: In
To prove: AE = BD
Proof:

In
AC = BC (given)     .....(1)

CE = CD (by CPCT)    .....(2)
Subtraction equation (2) from (1)

$⇒$AE = BD
â€‹Hence, proved.

#### Question 10:

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Given: $△\mathrm{ABC}$ in which AD is a median.
To prove: AB + AC > 2AD.
Proof:

Now, In
BD = DC ($\therefore$ AD is a median)
$\angle \mathrm{ADB}=\angle \mathrm{EDC}$ (vertically opposite angles)
$\therefore △\mathrm{ADB}\cong △\mathrm{EDC}$ (by SAS)
AB = EC (by CPCT)    .....(1)
In $△\mathrm{AEC},$ we have
AC + EC > AE      .....(2)
Using (1), (2) and (3), we get

#### Question 11:

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2(BD + AC)

Here, In ...............(1)
In ...........(2)
In ...............(3)
In ................(4)
Adding (1), (2), (3) and (4), we get
$⇒\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2\mathrm{OA}+2\mathrm{OB}+2\mathrm{OC}+2\mathrm{OD}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2\left(\mathrm{AO}+\mathrm{OC}\right)+2\left(\mathrm{BO}+\mathrm{OD}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2\left(\mathrm{AC}+\mathrm{BD}\right)$
Hence, verified.

#### Question 12:

Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

To prove: AB + BC + CD + DA > AC + BD
Proof:

We know, sum of any two sides of a triangle is greater than the third side
In $△\mathrm{ABC}$,
AB + BC > AC    .....(1)
In $△\mathrm{BCD},$
BC + CD > BD    .....(2)
In $△\mathrm{CDA}$
CD + DA > AC    .....(3)
In $△\mathrm{DAB},$
DA + AB > DB    .....(4)
Adding (1), (2), (3) and (4), we get
2(AB + BC + CD + DA) > 2(AC + BD0
⇒ AB + BC + CD + DA > AC + BC

Hence, proved.

#### Question 13:

In a triangle ABC, D is the mid-point of side AC such that BD = $\frac{1}{2}$ AC. Show that ∠ABC is a right angle.

Given: D is midpoint of AC in $△\mathrm{ABC}$ and BD = $\frac{1}{2}$ AC
To show: $\angle \mathrm{ABC}=90°$
Proof:

Here, BD = $\frac{1}{2}$AC (given)
$⇒$BD = AD = DC
In $△\mathrm{ABD}$,
$⇒\angle 1=\angle 2$     .....(1) (angle opposite to equal sides are equal)
Similarly, In $△\mathrm{BDC}$
BD = DC
$⇒\angle 3=\angle 4$     .....(2)
Let each angle be $x$.
$\therefore x+\left(x+x\right)+x=180°$
$⇒4x=180°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence,
$\begin{array}{rcl}\angle \mathrm{ABC}& =& x+x\\ & =& 45°+45°\\ & =& 90°\end{array}$

#### Question 14:

In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Given: In  and D is midpoint of AC.
To prove : BD = $\frac{1}{2}\mathrm{AC}$
Construction: Produce BD to E such that BD = DE and join E to C.
Proof:

In
$\angle \mathrm{ADB}=\angle \mathrm{CDE}$ (vertically opposite angle)
DB = DE (by construction)
$△\mathrm{ADB}\cong △\mathrm{CDE}$ (by SAS)
$\therefore \mathrm{AB}=\mathrm{EC}$ (by CPCT)
But are alternate angles
So, EC$\parallel$AB and BC is a transversal.
[cointerior angles]
$⇒90°+\angle \mathrm{BCE}=180°$ [$\therefore \angle \mathrm{ABC}=90°$
$⇒\angle \mathrm{BCE}=90°$
In
AB = EC (proved above)
BC = CB (common)
$\angle \mathrm{ABC}=\angle \mathrm{ECB}$ (each 90°)
$△\mathrm{ABC}\cong △\mathrm{ECB}$ (by SAS congruency rule)

Hence, proved.

#### Question 15:

Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.

Two lines $l$ and $m$ intersect at point O and P is a point on line $n$ such that P is equilateral from $l$ and $m$.
To prove: $n$ is bisector of angle formed by $l$ and $m$. i.e. $\angle \mathrm{POQ}=\angle \mathrm{POR}$
Proof:

In
PQ = PR (given)
OP = PO (common)
$\angle \mathrm{OQP}=\angle \mathrm{ORP}$ (each 90°)
$\therefore △\mathrm{OQP}\cong △\mathrm{ORP}$ (by RHS)
$\angle \mathrm{POQ}=\angle \mathrm{POR}$(by CPCT)
Hence, proved.

#### Question 16:

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.

Given: In trapezium ABCD, MN is perpendicular to both AB and CD. M and N are midpoints of AB and CD.
Proof: M and N are midpoints of AB and CD
⇒ AM = MB
And, DN = NC

In
AM = BM (given)
$\angle \mathrm{AMN}=\angle \mathrm{BMN}$ (each 90°)
MN = NM (common)
$△\mathrm{ANM}\cong △\mathrm{BNM}$ (by SAS)
$⇒\angle \mathrm{ANM}=\angle \mathrm{BNM}$ (by CPCT)

Now, in
AN = BN (From (2))

DN = NC (N is the midpoint of CD)
$△\mathrm{AND}\cong △\mathrm{BNC}$ (by SAS)

Hence, proved.

#### Question 17:

ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.

Given: ABCD is a quadrilateral such that AC bisects angle A and C
CB = CD
Proof:

In
$\angle \mathrm{DAC}=\angle \mathrm{BAC}$ (AC bisects $\angle \mathrm{A}$)
AC = AC (common)
$\angle \mathrm{DCA}=\angle \mathrm{BCA}$ (AC bisect $\angle \mathrm{C}$)
$△\mathrm{ADC}\cong △\mathrm{ABC}$ (by ASA congruency rule)
CB = CD (by CPCT)
Hence, proved.

#### Question 18:

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Given: $△\mathrm{ABC}$ is right triangle.
To prove: AC + AD = BC
Proof: Let AB = AC = $x$

By Pythagoras theorem, we have
BC = $\sqrt{{x}^{2}+{x}^{2}}=\sqrt{2}x$
By bisector theorem, we have
$\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BC}}{\mathrm{AC}}+1=\frac{\mathrm{BD}}{\mathrm{AD}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BC}+\mathrm{AC}}{\mathrm{AC}}=\frac{\mathrm{BD}+\mathrm{AD}}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{2}x+x}{x}=\frac{x}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{2}+1=\frac{x}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AD}=\frac{x}{\sqrt{2}+1}$
Now,
$\begin{array}{rcl}\therefore \mathrm{AC}+\mathrm{AD}& =& x+\frac{x}{\sqrt{2}+1}\\ & =& \frac{\sqrt{2}x\left(1+\sqrt{2}\right)}{\left(1+\sqrt{2}\right)}\\ & =& \sqrt{2}x\\ & =& \mathrm{BC}\end{array}$

Hence, proved.

#### Question 19:

AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of ∠B and ∠D decide which is greater.

Construction: Join BD

In $△\mathrm{BCD}$,
DC > BC ($\therefore$CD is largest side)
$\angle \mathrm{CBD}>\angle \mathrm{BDC}$ (angle opposite to greater side is greater)  .....(1)
In $△\mathrm{ABD}$
AD > AB (AB is smallest side)
$\angle \mathrm{ABD}>\angle \mathrm{ADB}$ (angle opposite to greater side is greater)  .....(2)
$\angle \mathrm{CBD}+\angle \mathrm{ABD}>\angle \mathrm{BDC}+\angle \mathrm{ADB}$
$\angle \mathrm{ABC}>\angle \mathrm{ADC}$
Hence, $\angle \mathrm{B}$ is  greater.

#### Question 20:

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Consider AC is the longest side in $△\mathrm{ABC}$.

So, AC > AB
$⇒\angle \mathrm{B}>\angle \mathrm{C}$ (angle opposite to greater side is greater)    .....(1)
Also, AC > BC
$\angle \mathrm{B}>\angle \mathrm{A}$ (angle opposite to greater side is greater)        .....(2)
Adding (1) and (2), we get
2$\angle \mathrm{B}>\angle \mathrm{A}+\angle \mathrm{C}$
Adding $\angle \mathrm{B}$ on both sides,

⇒ 3$\angle \mathrm{B}>180°$ ($\therefore$sum of angles of triangle = 180°)
$\angle \mathrm{B}>60°$
$⇒\mathrm{B}>\frac{2}{3}\left(90°\right)$
i.e. $\angle \mathrm{B}>\frac{2}{3}$ of the right angle.
Hence, proved.

#### Question 21:

ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.

To prove: AC is perpendicular bisector of BD.
Proof:

In
$△\mathrm{ABC}\cong △\mathrm{ADC}$ (by SSS)
$\angle \mathrm{BAO}=\angle \mathrm{DAO}$ ..........(AC bisects $\angle \mathrm{A}$)