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#### Question 1:

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is :
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

Given that, AD = 34 cm and AB = 30 cm. First, draw OL$\perp$ AB.
As, the perpendicular form the center of a circle to a chord bisects the chord. Now, $△\mathrm{OLA}$ is right angled, then
OA2 = OL2 + AL2                [by Pythagoras theorem]

Thus, the distance of the chord form the center is 8 cm.
Hence, the correct answer is option D.

#### Question 2:

In the figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to: (A) 2 cm
(B) 3 cm
(C) 4 cm
(D) 5 cm

The perpendicular from the center of a circle to a chord bisects the chord.
∴ AC = CB = $\frac{1}{2}$AB
= $\frac{1}{2}$$×$8
= 4 cm
It is given that OA = 5 cm.
$⇒{\mathrm{AO}}^{2}={\mathrm{AC}}^{2}+{\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(5\right)}^{2}={\left(4\right)}^{2}+{\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}⇒25=16+{\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{OC}}^{2}=25-16=9$
[∵Length is always positive, taking a positive square root]
Thus,
OA = OD = radius of a circle
OD = 5 cm
CD = OD − OC = 5 − 3 = 2 cm
Hence, the correct answer is option A.

#### Question 3:

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is :
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm

Given that, AB = 12 cm and BC = 16 cm.
In the circle, BC$\perp$AB. Thus, AC is the diameter of circle    [∵ diameter of a circle subtends a right angle to the circle] In right angled $△\mathrm{ABC}$, using Pythagoras theorem,

[taking positive square root, because diameter is always positive]

Thus, the radius of circle is 10 cm.
Hence, the correct answer is option C.

#### Question 4:

In the figure, if ∠ABC = 20°, then ∠AOC is equal to: (A) 20°
(B) 40°
(C) 60°
(D) 10°

Given that, ∠ABC = 20°.

The angle subtended by an arc at the centre is twice the angle subtended at the rest of the circle.
⇒ ∠AOC = 2∠ABC
⇒ ∠AOC = 2 × 20°
= 40°

Hence, the correct answer is option B.

#### Question 5:

In the figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to: (A) 30º
(B) 60º
(C) 90º
(D) 45º

The diameter subtends a right angle to the circle.

As, AC = BC
[∵ angles opposite to equal sides are equal] In $△\mathrm{ABC},$ by angle sum property of a triangle
$\angle \mathrm{CAB}+\angle \mathrm{ABC}+\angle \mathrm{BCA}=180°$

Using (1) and (2),
$⇒\angle \mathrm{CAB}+\angle \mathrm{CAB}+\angle 90°=180°$
$⇒2\angle \mathrm{CAB}=180°-90°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{CAB}=\frac{90°}{2}\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{CAB}=45°$
Hence, the correct answer is option D.

#### Question 6:

In the figure, if ∠OAB = 40°, then ∠ACB is equal to : (A) 50°
(B) 40°
(C) 60°
(D) 70°

In $△$OAB, OA = OB = radius of the circle

Also, by angle sum property of triangles in $△$AOB,
$\angle \mathrm{AOB}+\angle \mathrm{OBA}+\angle \mathrm{BAO}=180°$
$⇒\angle \mathrm{AOB}+40°+40°=180°$
$⇒\angle \mathrm{AOB}=180°–80°=100°$

The angle subtended by an arc at the centre of a circle is double the angle subtended by it at the rest of the circle.
$\angle \mathrm{AOB}=2\angle \mathrm{ACB}\phantom{\rule{0ex}{0ex}}⇒100°=2\angle \mathrm{ACB}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ACB}=\frac{100°}{2}=50°$
Hence, the correct answer is option A.

#### Question 7:

In the figure, if ∠DAB = 60°, ∠ABD = 50°, then ∠ACB is equal to: (A) 60°
(B) 50°
(C) 70°
(D) 80°

Given that, .
As,    [∵ angles in same segment of a circle are equal]

In $△\mathrm{ABD},$
[by angle sum property of a triangle]
$⇒50°+\angle \mathrm{ADB}+60°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ADB}=180°–110°=70°$
$⇒\angle \mathrm{ACB}=70°$

Hence, the correct answer is option C.

#### Question 8:

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to:
(A) 80°
(B) 50°
(C) 40°
(D) 30°

Given that, ABCD is a cyclic quadrilateral and . The sum of the opposite angles in a cyclic quadrilateral is 180°, then
$\angle \mathrm{ADC}+\angle \mathrm{ABC}=180°\phantom{\rule{0ex}{0ex}}⇒140°+\angle \mathrm{ABC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ABC}=180°-140°\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{ABC}=40°$

As,  AB is a diameter of the circle and $\angle \mathrm{ACB}$ is an angle in a semi-circle.
$\therefore \angle \mathrm{ACB}=90°$

In $△\mathrm{ABC},$ by angle sum property of a triangle
$\angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{ABC}=180°$
$⇒\angle \mathrm{BAC}+90°+40°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{BAC}=180°-130°=50°$
Hence, the correct answer is option B.

#### Question 9:

In the figure BC is a diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to : (A) 30º
(B) 45º
(C) 60º
(D) 120º

In $△\mathrm{AOB},$
[∵ angles opposite to equal sides are equal]

Now, angles in the same segment AC are equal.

$⇒\angle \mathrm{ADC}=60°$
Hence, the correct answer is option C.

#### Question 10:

In the figure, ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to: (A) 30°
(B) 45°
(C) 90°
(D) 60°

In $△$OAB, by angle sum property of a triangle,
$\angle \mathrm{OAB}+\angle \mathrm{ABO}+\angle \mathrm{BOA}=180°$
$\angle \mathrm{OAB}+\angle \mathrm{OAB}+90°=180°$        [∵ angles opposite to equal sides are equal]

Now, the angle subtended by an arc on the centre of the circle is twice the angle subtended by it at any other point on the circle.
$\therefore \angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ACB}=\frac{1}{2}×90°=45°$

In $△\mathrm{ACB}$,
$\angle \mathrm{ACB}+\angle \mathrm{CBA}+\angle \mathrm{CAB}=180°$
$⇒45°+30°+\angle \mathrm{CAB}=180°$
$⇒\angle \mathrm{CAB}=180°–75°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{CAB}=105°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{CAO}+\angle \mathrm{OAB}=105°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{CAO}+45°=105°$

$\therefore \angle \mathrm{CAO}=105°–45°=60°$
Hence, the correct answer is option D.

#### Question 1:

State whether the following statement is True or False.
Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then AB = CD.

The statement is true.
This is because the chords equidistant from the centre of circle are equal in length. #### Question 2:

State whether the following statement is True or False.
Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then ∠OAB = ∠OAC .

The statement is false.
In the figure below, AB and AC are two chords of a circle.
Join OB and OC. In
OA = OA  [common side]
OB = OC [radius of circle]

It cannot be seen that any angle or the third side of the two triangles is equal and thus, $△\mathrm{OAB}$ is not congruent to $△\mathrm{OAC}.$
$\therefore \angle \mathrm{OAB}\ne \angle \mathrm{OAC}.$

#### Question 3:

State whether the following statement is True or False.
Two congruent circles with centres O and O′ intersect at two points A and B. Then ∠AOB = ∠AO′B.

The statement is true.
Consider the figure given below: Join AB, OA and OB, O’A and BO’.

In
OA = AO’   [both circles have same radius]
OB = BO’   [both circles have same radius]
AB = AB   [common chord]

$⇒△\mathrm{AOB}=△\mathrm{AO}’\mathrm{B}$   [by SSS congruence rule]
Hence, $\angle \mathrm{AOB}=\angle \mathrm{AO}’\mathrm{B}$    [by CPCT]

#### Question 4:

State whether the following statement is True or False.
Through three collinear points, a circle can be drawn.

The statement is false.
This is because a circle can pass through only two collinear points, not three.

#### Question 5:

State whether the following statement is True or False.
A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.

The statement is true.

Consider the diameter of the circle as 6 cm.
⇒ AB = 6 cm
Then, radius of a circle =

#### Question 6:

State whether the following statement is True or False.
If AOB is a diameter of a circle and C is a point on the circle, then AC2 + BC2 = AB2.

The statement is true.

The diameter of a circle subtends a right angle at any other point on the circle.

If AOB is a diameter of a circle and C is a point on the circle, then $△$ACB is right angled at C.

In right angled $△$ACB, by Pythagoras theorem
AC2 + BC2 = AB2 #### Question 7:

State whether the following statement is True or False.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.

The statement is false.

The sum of opposite angles in a cyclic quadrilateral is 180°.
Now,
$\angle \mathrm{A}+\angle \mathrm{C}=90°+95°=185°\ne 180°$
and
$\angle \mathrm{B}+\angle \mathrm{D}=70°+105°=175°\ne 180°$

Here, the sum of opposite angles is not equal to 180°. Therefore, ABCD is not a cyclic quadrilateral.

#### Question 8:

State whether the following statement is True or False.
If A, B, C, D are four points such that ∠BAC = 30° and ∠BDC = 60°, then D is the centre of the circle through A, B and C.

The statement is false.

Consider the following case: Since, the angle subtended by an arc at the centre is double the angle subtended by it at any other point on the circle.
∠BAC = 30°

But now, Here, ∠BDC = 60°.
⇒ reflex(∠BDC) = 300°
Again, the angle subtended by an arc at the centre is double the angle subtended by it at any other point on the circle.
∠BAC = 150° ≠ 30°

#### Question 9:

State whether the following statement is True or False.
If A, B, C and D are four points such that ∠BAC = 45° and ∠BDC = 45°, then A, B, C, D are concyclic.

The statement is true.
As, Angles in the same segment of a circle are equal. As a result, A, B, C, and D are all concyclic.

#### Question 10:

State whether the following statement is True or False.
In the figure, if AOB is a diameter and ∠ADC = 120°, then ∠CAB = 30°. The statement is true.
Join CA and CB. As, ADCB is a cyclic quadrilateral. Thus, the sum of opposite angles of cyclic quadrilateral is 180°.
$\angle \mathrm{ADC}+\angle \mathrm{CBA}=180°$

In $△\mathrm{ACB}$, by angle sum property of a triangle],
$\angle \mathrm{CAB}+\angle \mathrm{CBA}+\angle \mathrm{ACB}=180°$
$⇒\angle \mathrm{CAB}+60°+90°=180°$           [∵ angle formed by the diameter to the circle is 90° i.e., $\angle \mathrm{ACB}=90°$]
$⇒\angle \mathrm{CAB}=180°–150°=30°$

#### Question 1:

If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

Let AXB and CYD be the two arcs of a circle arcs with centre O and radius r.

Now, when the two chords are obtained, i.e., AB and CD for the congruent arcs AXB and CYD, their repective chords will be equal.

Hence, AB : CD is equal to 1 : 1.

#### Question 2:

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.

Let PQ be the perpendicular bisector of AB, which intersects it at M and always passes through the centre of the circle O.
To prove: arc PXA ≅ arc PYB
Construction: Join AP and BP. In $△$APM and $△$BPM,
AM = MB      (∵ the perpendicular drawn from the centre of a circle to a chord, bisects it)
$\angle \mathrm{PMA}=\angle \mathrm{PMB}=90°$
PM = PM        (common)
Thus, $△\mathrm{APM}\cong △\mathrm{BPM}$   [by Side Angle Side Congruency]

∴ PA = PB

#### Question 3:

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.

Consider three non-collinear points on a circle, i.e., A, B and C.
Join AB, BC and CA.
Construction: Draw TU, PQ and RS, which are the perpendicular bisectors of AB, BC and CA respectively. To prove: The perpendicular bisectors are concurrent.
Proof: The points A, B and C are non-collinear.
Thus, the perpendicular bisectors TU, PQ and RS are not parallel and they will intersect at a point. Let the point of intersection be O.

Now, join OC, OA and OB.

Consider $△$AOU and $△$BOU.
AU = BU  (∵ TU is the perpendicular bisector of AB)
∠AUO = ∠BUO = 90°
OU = OU (common)
∴ $△\mathrm{AUO}\cong △\mathrm{BUO}$ by SAS congruence criterion.
⇒ OA = OB  (by CPCT)

Similarly, $△\mathrm{BUO}\cong △\mathrm{BPO}$ and $△\mathrm{BPO}\cong △\mathrm{CPO}$.
⇒ OA = OB = OC  (by CPCT)

Let OA = OB = OC = r.
This means that O is the only point equidistant from points on the circle A, B and C. Thus, O is the centre of the circle.
Therefore, the perpendicular bisectors of AB, BC and CA are concurrent.
Hence, proved.

#### Question 4:

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

Given: AB and AC are two equal chords whose centre is O
$\therefore \mathrm{AB}=\mathrm{AC}$
To prove: Center O lies on the bisector of $\angle \mathrm{BAC}.$
Construction: Join BC. Draw bisector AD of $\angle \mathrm{BAC}⇒\angle \mathrm{BAD}=\angle \mathrm{CAD}$ Proof: In ,
AB = AC       (given)
$\angle \mathrm{BAO}=\angle \mathrm{CAO}$  (by construction)
AO = AO       (common)
$\therefore △\mathrm{BAO}\cong △\mathrm{CAO}$       [by SAS congruence rule]
$⇒$BO = CO     [by CPCT]
and
$\angle \mathrm{BOA}=\angle \mathrm{COA}$       [by CPCT]

Also, BO = CO and $\angle \mathrm{BOA}=\angle \mathrm{COA}=90°$
Thus, AO is the perpendicular bisector of the chord BC.
Therefore, the bisector of $\angle \mathrm{BAC}$ i.e., AD passes through the centre O.
Hence proved.

#### Question 5:

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.

Given that, AB and CD are two chords of a circle with centre O. PQ is the diameter of the circle bisecting the chord AB and CD at points L and M respectively. To prove: AB $\parallel$ CD

Proof:
Now, L is the mid-point of AB. The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
$\therefore \mathrm{OL}\perp \mathrm{AB}$.

Also, OM$\perp$CD.

From (1) and (2),
$\angle \mathrm{ALO}=\angle \mathrm{OMD}=90°$

Since, these are alternating angles.
Therefore, AB $\parallel$ CD.
Hence, proved.

#### Question 6:

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$.

In a circle, consider a quadrilateral ABCD with centre A.
To prove: $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$
Construction: Join AC and BD. Proof: In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
As, arc CD subtends $\angle \mathrm{DAC}$ at the centre and $\angle \mathrm{CBD}$ at a point B.

Again, arc BC subtends $\angle \mathrm{CAB}$ at the centre and $\angle \mathrm{CDB}$ at a point D in the remaining part of the circle.

On adding (1) and (2),
$\angle \mathrm{DAC}+\angle \mathrm{CAB}=2\angle \mathrm{CBD}+2\angle \mathrm{CDB}$
$⇒\angle \mathrm{BAD}=2\left(\angle \mathrm{CBD}+\angle \mathrm{CDB}\right)$
$⇒\angle \mathrm{CDB}+\angle \mathrm{CBD}=\frac{1}{2}\angle \mathrm{BAD}$
Hence, proved.

#### Question 7:

O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.

Consider a triangle $△$ABC inside a circle with centre O. Also, D is the mid-point of BC.
To prove:
Construction: Join OB, OD and OC. Proof: In $△$BOD and $△$COD,
OB = OC      [radius of circle]
BD = DC      [D is the mid-point of BC]
Also, OD = OD      [Common]
$\therefore △\mathrm{BOD}\cong △\mathrm{COD}$     [by SSS congruence rule]
Then, $\angle \mathrm{BOD}=\angle \mathrm{COD}$ [CPCT]        .....(1)

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore 2\angle \mathrm{BAC}=\angle \mathrm{BOC}$
$⇒\angle \mathrm{BAC}=\frac{2}{2}\angle \mathrm{BOD}$   [As $\angle \mathrm{BOC}=2\angle \mathrm{BOD}\right]$ [Using (1)]
$⇒\angle \mathrm{BAC}=\angle \mathrm{BOD}$

Hence, proved.

#### Question 8:

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.

Given that, two right angled triangles  have a common hypotenuse AB.
To prove: $\angle \mathrm{BAC}=\angle \mathrm{BDC}$
Construction: Join CD. Proof: Since,  are right angled triangles.
Therefore,
∠C + ∠D = 90° + 90°
⇒ ∠C + ∠D = 180°
Thus, ADBC is a cyclic quadrilateral as sum of opposite angles of a cyclic quadrilateral is 180°.

Also, ∠BAC and ∠BDC lie in the same segment BC and angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC.

Hence, proved.

#### Question 9:

Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.

In $△\mathrm{BOA},$ OB = OA  [radius of circle]
[∵ angles opposite to equal sides are equal] In $△$OAB, by angle sum property of a triangle,
$\angle \mathrm{OBA}+\angle \mathrm{OAB}+\angle \mathrm{AOB}=180°$

In  [radius of a circle]
[∵ angles opposite to equal sides are equal]

Now, by angle sum property of a triangle,
$\angle \mathrm{AOC}+\angle \mathrm{OAC}+\angle \mathrm{OCA}=180°$

Hence, $\angle \mathrm{BAC}$ is 60°.

#### Question 10:

If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

In ΔABC,
.
To prove: Points B, C, M and N are concyclic.
Construction: Construct a circle passing through the points B, C, M and N. Proof:
Consider BC as the diameter of the circle.
Then, BC subtends a 90° at any other point on the circle.
Here, ∠BNC = 90°.
Thus, point N lies on the circle.

Similarly, the point M also lies on the circle.
Therefore, BCMN form a concyclic quadrilateral.
Hence, proved.

#### Question 11:

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

$△$ABC is an isosceles triangle such that AB = AC and DE ∥ BC.
To prove: Quadrilateral BCDE is a cyclic quadrilateral.
Construction: Draw a circle that passes through the points B, C, D and E. Proof: In $△\mathrm{ABC}$,
AB = AC         [∵ equal sides of an isosceles triangle]
[angle opposite to the equal sides are equal]

As, DE ∥ BD.
[corresponding angles]

On adding $\angle \mathrm{EDC}$ on both sides in (2),
$\angle \mathrm{ADE}+\angle \mathrm{EDC}=\angle \mathrm{ACB}+\angle \mathrm{EDC}$
$⇒180°=\angle \mathrm{ACB}+\angle \mathrm{EDC}$           [∵ from linear pair axiom]
$⇒\angle \mathrm{EDC}+\angle \mathrm{ABC}=180°$
Hence, as the sum of opposite angle is 180°. BCDE is a cyclic quadrilateral.

#### Question 12:

If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

Let ABCD be a cyclic quadrilateral where AD = BC.
Construction: Join AC and BD. To prove: AC = BD
Proof: In $△$AOD and $△$BOC,
[∵ angles subtended in the same segment are equal]
Also, AD = BC.
$\therefore △\mathrm{AOD}\cong △\mathrm{BOC}$      [by ASA congruence rule]

By CPCT,
AO = OB           .....(1)
OC = OD           .....(2)

Adding (1) and (2),
AO + OC = OB + OD
⇒ AC = BD
Hence, proved.

#### Question 13:

The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.

A circle is circumscribed on a ΔABC having centre O.
To prove: $\angle \mathrm{OBC}+\angle \mathrm{BAC}=90°$
Construction: Join BO and CO. Proof : Let $\angle \mathrm{OBC}=\angle \mathrm{OCB}=\mathrm{\theta }$.
In $△\mathrm{OBC}$, by angle sum property of triangles,

$⇒\angle \mathrm{BOC}+\mathrm{\theta }+\mathrm{\theta }=180°$
$⇒\angle \mathrm{BOC}=180°-2\mathrm{\theta }$

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Hence, proved.

#### Question 14:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.

As, AB is a chord of a circle, which is equal to the radius of the circle.
⇒ AB = BO
Construction: Join AC, OA and BC. Since, OA = OB = Radius of circle
$⇒$OA = AB = BO
Thus, $△\mathrm{OAB}$ is an equilateral triangle.
$⇒\angle \mathrm{AOB}=60°$         [All angles of an equilateral triangle is 60°]

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at remaining part of the circle.
$⇒\angle \mathrm{AOB}=2\angle \mathrm{ACB}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ACB}=\frac{60°}{2}=30°$
Hence, the angle subtended by this chord at a point in major segment is 30°.

#### Question 15:

In the figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE. Given: $\angle \mathrm{ADC}=130°$ and chord BC = chord BE.
Let the points A, B, C and D form a cyclic quadrilateral.

As, the sum of opposite angles of a cyclic quadrilateral $△$DCB is 180°.
Then, $\angle \mathrm{ADC}+\angle \mathrm{OBC}=180°$
$⇒130°+\angle \mathrm{OBC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{OBC}=180°-130°=50°$

In $△$BOC and $△$BOE,
BC = BE [given that the chords are equal]
∠BCO = ∠BEO       [angles opposite to equal sides are equal)
and OB = OB [common side]
⇒ $△$BOC $\cong$$△$BOE

$⇒\angle \mathrm{CBE}=\angle \mathrm{CBO}+\angle \mathrm{EBO}\phantom{\rule{0ex}{0ex}}=50°+50°\phantom{\rule{0ex}{0ex}}=100°$

Hence, $\angle \mathrm{CBE}=100°$.

#### Question 16:

In the figure, ∠ACB = 40º. Find ∠OAB. Given that, $\angle \mathrm{ACB}=40°$.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

In $△\mathrm{AOB}$,
AO = OB [radius of a circle]
[∵ angles opposite to the equal sides of a triangle are equal]

Also, the sum of all three angles in a triangle AOB is 180°.

Hence, $\angle \mathrm{OAB}$ is 50°.

#### Question 17:

A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130°. Find ∠BAC.

Given that, quadrilateral ABCD is inscribed in a circle having centre O. Also, $\angle \mathrm{ADC}=130°$. As, ABCD is a quadrilateral inscribed in a circle, therefore ABCD become a cyclic quadrilateral.

Therefore, the sum of opposite angles of a cyclic quadrilateral is 180°.
Then,
$\angle \mathrm{ADC}+\angle \mathrm{ABC}=180°$
$⇒130°+\angle \mathrm{ABC}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ABC}=50°$

Now, AB is a diameter of a circle. Thus, the angle subtended by AB at any point of the circle is right angle.
$\therefore \angle \mathrm{ACB}=90°$

In $△\mathrm{ABC}$, by angle sum property of a triangle,
$\angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{ABC}=180°$
$⇒\angle \mathrm{BAC}+90°+50°=180°$
$\begin{array}{rcl}& ⇒& \angle \mathrm{BAC}=180°-\left(90°+50°\right)\\ & =& 180°-140°\\ & =& 40°\end{array}$

Hence, $\angle \mathrm{BAC}$ is 40°.

#### Question 18:

Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2OO′.

Given, two circles having centres O and O′ intersect at points A and 8.
Construction: Join OO, OP, O′Q, OM and O′N. Draw line PQ parallel to OO′. To prove: PQ = 2OO′
Proof:
In $△\mathrm{OPB}$,
BM = MP        [OM is the perpendicular bisector of PB]
and
in $△$O′BQ,
BN = NQ         [O′ N is the perpendicular bisector of BQ]
$\therefore \mathrm{BM}+\mathrm{BN}=\mathrm{PM}+\mathrm{NQ}$

Adding (BM + BN) on both the sides,
$⇒2\left(\mathrm{BM}+\mathrm{BN}\right)=\mathrm{BM}+\mathrm{BN}+\mathrm{PM}+\mathrm{NQ}$
⇒ 2OO′ = (BM + MP) + (BN + NQ)        [∵ OO′ = MN = MB + BN]
= BP + BQ
= PQ
​⇒ 2OO′ = PQ
Hence, proved.

#### Question 19:

In the figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED. Join AE. Write D
Since, A, C, D and E are four point on a circle, therefore ACDE is a cyclic quadrilateral.
$⇒$          [sum of opposite angles in a cyclic quadrilateral is 180°]        .....(1)
Now, the diameter of a circle subtends a right angle at any point on the circle.

On adding (1) and (2),
$\left(\angle \mathrm{ACD}+\angle \mathrm{AED}\right)+\angle \mathrm{AEB}=180°+90°=270°$
$⇒\angle \mathrm{ACD}+\angle \mathrm{BED}=270°$

Hence, $\angle \mathrm{ACD}+\angle \mathrm{BED}$ is 270°.

#### Question 20:

In the figure, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC. Given that, .
In $△\mathrm{AOB}$.
OA = OB    [radius of a circle]
$⇒\angle \mathrm{OBA}=\angle \mathrm{BAO}=30°$  [angles opposite to equal sides are equal]

In $△\mathrm{AOB}$, by angle sum property of a triangle,
$⇒\angle \mathrm{AOB}+\angle \mathrm{OBA}+\angle \mathrm{BAO}=180°$
$⇒\angle \mathrm{AOB}+30°+30°=180°$

Now, in $△\mathrm{OCB},$
OC = OB   [radius of a circle]
$⇒\angle \mathrm{OBC}=\angle \mathrm{OCB}=57°$     [angles opposite to equal sides are equal]

In $△\mathrm{OCB},$ by angle sum property of a triangle,
$\angle \mathrm{COB}+\angle \mathrm{OCB}+\angle \mathrm{CBO}=180°$
$\therefore \angle \mathrm{COB}=180°-\left(\angle \mathrm{OCB}+\angle \mathrm{OBC}\right)$

$⇒\angle \mathrm{COB}=\angle \mathrm{BOC}=66°$

From (1),
$\angle \mathrm{AOB}=120°$.

Hence, $\angle \mathrm{AOC}$ is 54° and $\angle \mathrm{BOC}$ is 66°.

#### Question 1:

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Given: Let AB and CD are two equal chords of a circle meeting at point E.
To prove: AE = CE and BE = DE
Construction: Draw OM$\perp$AB, ON$\perp$CD and join OE, where O is the centre of circle. Proof: In
OM = ON      [equal chords are equidistant from the centre]
OE = OE       [common side]
Also, $\angle \mathrm{OME}=\mathrm{ONE}$  [each 90°]

$⇒\mathrm{EM}=\mathrm{EN}$  [CPCT]      .....(1)

Also, AB = CD.
On dividing both sides by 2,

[∵Perpendicular drawn from the centre of acircle to a chord, bisects it]

On adding (1) and (2),
EM + AM = EN + CN

Now, AB = CD.

On subtracting AE from both the sides,
AB $-$ AE = CD $-$ AE
⇒ BE = CD $-$ CE    [from (3)]
⇒ BE = DE
Hence, proved.

#### Question 2:

If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Given: ABCD is a trapezium with equal non-parallel sides AD and BC.
To prove: Trapezium ABCD is a cyclic.
Construction: Join BE such that BE$\parallel$AD. Proof: As, AB$\parallel$DE and AD$\parallel$BE
Then, the quadrilateral ABED is a parallelogram.
[opposite angles of a parallelogram are equal]
and AD = BE       .....(2)   [opposite sides of a parallelogram are equal]
But AD = BC       .....(3)

From (2) and (3),
BC = BE
$⇒\angle \mathrm{BEC}=\angle \mathrm{BCE}$     .....(4)      [angles opposite to equal sides are equal]

Also,
$\angle \mathrm{BEC}+\angle \mathrm{BED}=180°$        [linear pair axiom]
$\therefore \angle \mathrm{BCE}+\angle \mathrm{BAD}=180°$    [from (1) and (4)]

In a cyclic quadrilateral, the sum of opposite angles is 180°.
Thus, trapezium ABCD is a cyclic.
Hence, proved.

#### Question 3:

If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.

In $△$ABC, R and Q are mid points of AB and CA respectively.
By mid-point theorem, RQ ∥ BC.
Similarly, PQ ∥ AB and PR ∥ CA.

In quadrilateral BPQR, BP ∥ RQ and PQ ∥ BR. Thus, the quadrilateral BPQR is a parallelogram. Similarly, quadrilateral ARPQ is a parallelogram.
∴ ∠A = ∠RPQ      (∵ Opposite sides of parallelogram are equal)

Also, PR ∥ AC and PC is the transversal.
∴ ∠BPR = ∠C      (Corresponding angles)
∠DPQ = ∠DPR + ∠RPQ
= ∠A + ∠C        .....(1)

And RQ ∥ BC and BR is the transversal.
∴ ∠ARO = ∠B         (Corresponding angles)      .....(2)

In $△$ABD, R is the mid point of AB and OR ∥ BD.
∴ O is the mid point of AD (Converse of mid point theorem)
⇒ OA = OD

In $△$AOR and $△$DOR,
OA = OD
∠AOR = ∠DOR = 90°     {∠ROD = ∠ODP (Alternate angles) & ∠AOR = ∠ROD = 90° (linear pair)}
OR = OR (Common)
$△$AOR $\cong$$△$DOR        (by SAS congruence criterion)

⇒ ∠ARO = ∠DRO (CPCT)
⇒ ∠DRO = ∠B                (Using (2))

∠DRO + ∠DPQ = ∠B + ( ∠A + ∠C)
= ∠A + ∠B + ∠C       (Using (1))
⇒ ∠DRO + ∠DPQ = 180°                    (∵ ∠A + ∠B + ∠C = 180°)

Therefore, quadrilateral PRQD is a cyclic quadrilateral. Thus, the points P, Q, R and D are concyclic.
Hence, proved.

#### Question 4:

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.

Given: ABCD is a parallelogram.
Let a circle whose centre is O passes through A and B such that it intersects AD at P and BC at Q.
To prove: Points P, Q, C and D are concyclic.
Construction: Join point P to Q. Thus, PQ line segment is constructed. Proof:
As,        [∵ exterior angle property of cyclic quadrilateral]
But $\angle \mathrm{A}=\angle \mathrm{C}$       [∵ opposite angles of a parallelogram are equal]
∴ $\angle 1=\angle \mathrm{C}$            .....(1)

But
$\angle \mathrm{C}+\angle \mathrm{D}=180°$    [On the same side of the transversal, the sum of co-interior angles is 180°]
$⇒\angle 1+\angle \mathrm{D}=180°$      [from (1)]
Therefore, the quadrilateral QCDP is cyclic.
So, the points P, Q, C and D are concyclic.
Hence, proved.

#### Question 5:

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

Given: $△\mathrm{ABC}$ is inscribed in a circle. The bisector of $\angle \mathrm{A}$ and perpendicular bisector of BC intersect at point O.
To prove: The angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of $△$ABC.
Construction: Join BO and OC. Proof: Let the angle bisector of ∠A intersect the circumcircle of $△$ABC at point D. Now, join DC and DB.

Since ,angles in the same segment are equal.
∴ ∠BCD = ∠BAD
⇒ ∠BCD = ∠BAD = $\frac{1}{2}$∠A          .....(1)

Similarly,
∠DBC = ∠DAC = $\frac{1}{2}$∠A          .....(2)

From (1) and (2),
∠DBC = ∠BCD
⇒ BD = DC        [∵ sides opposite to equal angles are equal]
Thus, D lies on the perpendicular bisector of BC.
Therefore, the angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of $△$ABC.
Hence, proved.

#### Question 6:

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle. In a circle AYDZBWCX, two chords AB and CD intersect at right angles.
To prove: arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle.
Construction: Draw diameter EF parallel to CD with M as the centre. Proof: As, CD$\parallel$EF.
⇒ arc EC = arc FD                  .....(1)
arc ECXA = arc EWB            [symmetrical in relation to a circle's diameter]
⇒ arc AF = arc BF                  .....(2)

Now, arc ECXAYDF = Semi-circle
⇒ arc EA + arc AF = Semi-circle
⇒ arc EC + arc CXA + arc BF = Semi-circle  [from (2)]
⇒ arc DF + arc CXA + arc BF = Semi-circle  [from (1)]
⇒ arc DF + arc FB + arc CXA = Semi-circle
⇒ arc DZB + arc CXA = Semi-circle
As, the circle divides itself in two semi-circles. Therefore, the remaining portion of the circle is also equal to the semi-circle.
Thus, arc AYD + arc BWC = Semi-circle
Hence, proved.

#### Question 7:

If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of ∠BPC.

Given: P is any point on the minor arc BC which does not coincide with B or C and $△$ABC is an equilateral triangle inscribed on a circle.
To prove: PA is an angle bisector of $\angle \mathrm{BPC}$.
Construction: Join PB and PC. Proof: As, $△\mathrm{ABC}$ is an equilateral triangle.
$\angle 3=\angle 4=60°$

Also,
[∵ angles in the same segment AB]
[∵ angles in the same segment AC]
$\therefore \angle 1=\angle 2=60°$

Thus, PA is the bisector of $\angle \mathrm{BPC}.$
Hence, proved.

#### Question 8:

In the figure AB and CD are two chords of a circle intersecting each other at point E. Prove that $\angle \mathrm{AEC}=\frac{1}{2}$ (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre). Given, two chords AB and CD intersecting each other at point E.
To prove: $\angle \mathrm{AEC}=\frac{1}{2}$ [angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre]
Construction: Extend the line DO and BO at the points I and H on the circle. Then, join AC. Proof: In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it on the remaining part of the circle.

In $△$AOC,
OC = OA  [radius of circle]
$\angle \mathrm{OCA}=\angle 4$  [∵ angles opposite to equal sides are equal]

Also, by angle sum property of triangles,
$\angle \mathrm{AOC}+\angle \mathrm{OCA}+\angle 4=180°$

In $△\mathrm{OCD}$, angles opposite to equal sides are equal.
$\therefore \angle 6=\angle \mathrm{ECO}$

Then, in $△\mathrm{AEC}$, by angle property sum of a triangle,
$\angle \mathrm{AEC}+\angle \mathrm{ECA}+\angle \mathrm{CAE}=180°$
$\begin{array}{rcl}& ⇒& \angle \mathrm{AEC}=180°-\left(\angle \mathrm{ECA}+\angle \mathrm{CAE}\right)\\ & ⇒& \angle \mathrm{AEC}=180°-\left[\left(\angle \mathrm{ECO}+\angle \mathrm{OCA}\right)+\angle \mathrm{CAO}+\angle \mathrm{OAE}\right]\\ & =& 180°-\left(\angle 6+\angle 4+\angle 4+\angle 5\right)\end{array}$
$\begin{array}{rcl}& ⇒& \angle \mathrm{AEC}=180°-\left(2\angle 4+\angle 5+\angle 6\right)\\ & =& 180°-\left(180°-\angle \mathrm{AOC}+\angle 7+\angle 6\right)\end{array}$
[from (3) and in  as angles opposite to equal sides are equal.

$⇒\angle \mathrm{AEC}=\frac{1}{2}\left(\angle \mathrm{AOC}+\angle \mathrm{DOB}\right)$
$=\frac{1}{2}$ [angle subtended by arc CXA at the center + angle subtended by arc DYB at the center]
Hence, proved.

#### Question 9:

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

Given that, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B.
To prove: PQ is the diameter of the circle.
Construction: Join QD and QC. Proof: ABCD is a cyclic quadrilateral. The bisectors of opposite angles of the cyclic quadrilateral, ∠A and ∠C, intersect the circle circumscribing at the points P and Q respectively.
Now, the opposite angles of a cyclic quadrilateral are supplementary.
⇒ ∠A + ∠C = 180°
$⇒\frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{C}=\frac{1}{2}×180°=90°$
⇒ ∠PAB + ∠BCQ = 90°

But ∠BCQ = ∠BAQ             [∵ Angles in the segment of a circle are equal]
∴ ∠PAB + ∠BAQ = 90°
⇒ ∠PAQ = 90°
Thus, ∠PAQ is in semicircle.
∴ PQ is diameter of the circle.
Hence, proved.

#### Question 10:

A circle has radius $\sqrt{2}$ cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45º.

Given, a circle having centre O with Ab = 2 cm be a chord of a circle. The radius OM is divided by chord AB in two equal segments.
To prove: $\angle \mathrm{APB}=45°$ Proof: As, the perpendiculard rawn from the centre of a circle to a chord, bisects it.
∴ AN = NB = 1 cm
and OB =

In $△\mathrm{ONB}$, by Pythagoras theorem,
${\mathrm{OB}}^{2}={\mathrm{ON}}^{2}+{\mathrm{NB}}^{2}$

[using positive square root, as distance is always positive]

Also,
$\angle \mathrm{ONB}=90°$        [∵ ON is the perpendicular bisector of the chord AB]
$\therefore \angle \mathrm{NOB}=\angle \mathrm{NBO}=45°$

Similarly, $\angle \mathrm{AON}=45°$.

Then, $\angle \mathrm{AOB}=\angle \mathrm{AON}+\angle \mathrm{NOB}$
= 45° + 45°
= 90°

The angle subtended by an arc at the centre of the circle if twice the angle subtended by it at any other point on the circle.
$\begin{array}{rcl}\therefore \angle \mathrm{APB}& =& \frac{1}{2}\angle \mathrm{AOB}\\ & =& \frac{90°}{2}\\ & =& 45°\end{array}$
Hence, proved.

#### Question 11:

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.

Given: Two equal chords AB and CD of a circle intersecting at a point P.
To prove: PB = PD
Construction: Join OP, draw OL$\perp$AB and OM$\perp$CD. Proof: As, AB = CD.
⇒ OL = OM      [∵ equal chords are equidistant from the centre]

In ,
OL = OM
$\angle \mathrm{OLP}=\angle \mathrm{OMP}$    [right angle i.e. 90°]
and OP = OP       [common side]
$\therefore △\mathrm{OLP}\cong △\mathrm{OMP}$      [by RHS congruence rule]
⇒ LP = MP       [by CPCT]    .....(1)

Now, AB = CD.
$⇒\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{CD}$  [dividing by 2 on both sides]

[perpendicular draw from centre to the circle bisects the chord ⇒ AL = LB and CM = MD]

On subtracting , we get
LP $-$ BL = MP $-$ DM
$⇒$PB = PD
Hence, proved.

#### Question 12:

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q2 = p2 + 3r2.

Given: In a circle of radius r, there are two chords AB and AC such that AB = 2AC. Then, the distance of AB and  AC from the centre are p and q respectively.
To prove: 4qp3r2,
Proof: Let AC = a, then AB = 2a. From the centre O, perpendiculars are drawn to the chords AC and AB which intersect them at points M and N respectively.
$\therefore \mathrm{AM}=\mathrm{MC}=\frac{a}{2}$ and AN = NB = a

In $△\mathrm{OAM}$, by Pythagoras theorem,
${\mathrm{AO}}^{2}={\mathrm{AM}}^{2}+{\mathrm{MO}}^{2}$

In $△\mathrm{OAN}$, by Pythagoras theorem,
${\mathrm{AO}}^{2}={\left(\mathrm{AN}\right)}^{2}+{\left(\mathrm{NO}\right)}^{2}$

From (1) and (2),
${\left(\frac{a}{2}\right)}^{2}+{q}^{2}={a}^{2}+{p}^{2}$
$⇒\frac{{a}^{2}}{4}+{q}^{2}={a}^{2}+{p}^{2}$
$⇒{a}^{2}+4{q}^{2}=4{a}^{2}+4{p}^{2}$
$⇒4{q}^{2}=3{a}^{2}+4{p}^{2}\phantom{\rule{0ex}{0ex}}⇒4{q}^{2}={p}^{2}+3\left({a}^{2}+{p}^{2}\right)$
$⇒4{q}^{2}={p}^{2}+3{r}^{3}$           [In right angled triangle ]
Hence, proved.

#### Question 13:

In the figure, O is the centre of the circle, ∠BCO = 30°. Find x and y. Given, O is the centre of the circle and $\angle \mathrm{BCO}=30°$. Join OB and AC and let the perpendicular from the centre of the circle intersect the chord BC at E. In $△\mathrm{BOC},$
CO = BO  [radius of circle]
$\therefore \angle \mathrm{OBC}=\angle \mathrm{OCB}=30°$  [∵ Angles opposite to equal sides are equal]

By applying angle sum property of a triangle,
$\begin{array}{rcl}& ⇒& \angle \mathrm{BOC}=180°-\left(\angle \mathrm{OBC}+\angle \mathrm{OCE}\right)\\ & =& 180°-\left(30°+30°\right)\\ & =& 120°\end{array}$

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$⇒\angle \mathrm{BOC}=2\angle \mathrm{BAC}$
$\therefore \angle \mathrm{BAC}=\frac{120°}{2}=60°$

Now, consider $△$ABE and $△$ACE.
AE = AE       (common)
$\angle \mathrm{AEB}=\angle \mathrm{AEC}=90°$
BE = EC        (the perpendicular from the centre of the circle to a chord, bisects it)
$\therefore △$ABE ACE     (by SAS congruence)
[AE is an angle bisector of angle A]
$⇒x=30°$

In $△$ABE, by angle sum property of a triangle,
$\angle \mathrm{BAE}+\angle \mathrm{EBA}+\angle \mathrm{AEB}=180°$
$⇒30°+\angle \mathrm{EBA}+90°=180°$
$\begin{array}{rcl}\therefore \angle \mathrm{EBA}& =& 180°-\left(90°+30°\right)\\ & =& 180°-120°\\ & =& 60°\end{array}$

Thus, $\angle \mathrm{EBA}=60°$.

$⇒45°+y=60°\phantom{\rule{0ex}{0ex}}⇒y=60°-45°\phantom{\rule{0ex}{0ex}}\therefore y=15°$
Hence, x is 30° and y is 15°.

#### Question 14:

In the figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB. In the figure below, BD = OD, CD$\perp$AB
In $△$OBD,
BD = OD
and OD = OB  [radius of circle]
$⇒\mathrm{OB}=\mathrm{OD}=\mathrm{BD}$
Then, $△$ODB is an equilateral triangle.
$⇒\angle \mathrm{BOD}=\angle \mathrm{OBD}=\angle \mathrm{ODB}=60°$ In ,
MB = MB          [common side]
$\angle \mathrm{CMB}=\angle \mathrm{BMD}=90°$
CM = MD         (∵ In a circle, the perpendicular drawn to a chord from the centre of the circle, bisects it)
$\therefore △\mathrm{MBC}\cong △\mathrm{MBD}$     [by SAS congruence rule]
$⇒\angle \mathrm{MBC}=\angle \mathrm{MBD}$     [by CPCT]

As, AB is a diameter of the circle. Therefore, angle subtended by it at any point on the circle is a right angle.
$\therefore \angle \mathrm{ACB}=90°$

In $△$ACB, by angle sum property of a triangle,
$\angle \mathrm{CAB}+\angle \mathrm{CBA}+\angle \mathrm{ACB}=180°$
$⇒\angle \mathrm{CAB}+60°+90°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{CAB}=180°-\left(60°+90°\right)=30°$
Hence, the value of $\angle \mathrm{CAB}$ is 30°.

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