## Chapter 11 Circles Ex 11.5

**Question 1.** **In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.**

**Solution:**

**Question 2.** **A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.** **Solution:**

Let BC be chord, which is equal to the radius. Join OB and OC.

**Question 3.** **In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.**

**Solution:**

**Question 4.** **In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.**

**Solution:**

∵ ∠BDC = ∠BAC …(i)

(Since, the angles in the same segment are equal)

Now , in ∆ABC

∴ ∠A + ∠B+ ∠C= 180°

⇒ ∠A+ 69°+ 31° = 180°

⇒ ∠A + 100° = 180°

∴ ∠A = 180° – 100° = 80°

⇒ ∠BAC=80°

∴ From Eq.(i)∠BDC = 80°

**Question 5.** **In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.**

**Solution:**

∴ ∠AEB = 180° – 130° = 50° (Linear Pair) …(i)

⇒ ∠CED = ∠AEB = 50° (Vertically opposite)

Again ∠ABD = ∠ACD (Since, the angles in the same segment are equal)

∠ABE = ∠ECD

⇒ ∠ABE = 180° …(ii)

∴ In ∆ CDE

∠A+ 20° + 50° = 180° [From Eqs. (i) and (ii)]

∠A + 70° = 180°

∴ ∠A = 180°- 70° =110°

Hence ∠BAC = 110°

**Question 6.** **ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.** **Solution:**

Angles in the same segment are equal.

∴ ∠BDC = ∠BAC

∴ ∠BDC = 30°

In ∆ BCD, we have

∴ ∠BDC + ∠DBC + ∠BCD = 180° (Given, ∠DBC = 70° and ∠BDC = 30°)

∴ 30° + 70° + ∠BCD = 180°

∴ ∠BCD= 180°-30°-70° = 80°

If AB = BC, then ∠BCA = ∠BAC= 80° (Angles opposite to equal sides in a triangle are equal)

Now, ∠ECD = ∠BCD – ∠BCA = 80° – 30P = 50° (∵ ∠BCD = 80° and ∠BCA =30°)

Hence, ∠BCD = 80°

and ∠ECD = 50°

**Question 7.** **If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.** **Solution:** **Given:** Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.

**Question 8.** **If the non-parallel sides of a trapezium are equal, prove that it is cyclic.** **Solution:** **Given:** Non-parallel sides PS and QR of a trapezium PQRS are equal.

**To prove:** ABCD is a cyclic trapezium. **Construction:** Draw SM ⊥ PQ and RN ⊥ PQ.

Proof In ∆SMP and ∆RNQ, we get

SP = RQ (Given)

∠SMP = ∠RNQ (Each = 90°)

and SM = RN

(∵ Distance between two parallel lines is always equal)

∴ By RHS criterion, we get

∆ SMP ≅ ∆ RNQ

So, ∠P = ∠Q (By CPCT)

and ∠PSM = ∠QRN

Now, ∠PSM = ∠QRN

∴ 90° + ∠PSM = 90° + ∠QRN (Adding both sides 90°)

∴ ∠MSR + ∠PSM = ∠NRS + ∠QRN (∵∠MSR = ∠NRS = 90°)

So, ∠PSR = ∠QRS

i.e., ∠S = ∠R

Thus, ∠P = ∠Q and ∠R = ∠S …(i)

∴ ∠P+ ∠Q+ ∠R+ ∠S = 360° (∵ Sum of the angles of a quadrilateral is 360°)

∴ 2∠S + ∠Q = 360° [From Eq. (i)]

∠S+∠O = 180°

Hence, PQRS is a cyclic trape∠ium.

**Question 9.** **Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A,D and P, Q respectively (see figure). Prove that ∠ ACP = ∠QCD.**

**Solution:** **Given:** Two circles intersect at two points B and C. Through B two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q, respectively. **To prove:** ∠ACP = ∠QCD **Proof:** In circle I, ∠ACP = ∠ABP (Angles in the same segment) …(i)

In circle II, ∠QCD = ∠QBD{Angles in the same segment)…(ii)

∠ABP = ∠QBD (Vertically opposite angles)

From Eqs. (i) and (ii), we get ∠ACP = ∠QCD

**Question 10.** **If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.** **Solution:** **Given:** Two circles are drawn with sides AC and AB of AABC as diameters . Both circles intersect each other at D. **To prove:** D lies on BC. **Construction:** Join AD. **Proof:** Since, AC and AB are the diameters of the two circles.

∠ADB = 90° ( ∴ Angles in a semi-circle) …(i)

and ∠ADC = 90° (Angles in a semi-circle) …(ii)

On adding Eqs. (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90° = 180°

Hence, BCD is a straight line.

So, D lies on BC.

**Question 11.** **ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.** **Solution:**

Since, ∆ADC and ∆ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.

∵ Angles in the same segment are equal.

∴ ∠CBD = ∠CAD

**Question 12.** **Prove that a cyclic parallelogram is a rectangle.** **Solution:** **Given:** PQRS is a parallelogram inscribed in a circle. **To prove**: PQRS is a rectangle.

**Proof:** Since, PQRS is a cyclic quadrilateral.

∴ ∠P+∠R = 180°

(∵ Sum of opposite angles in a cyclic quadrilateral is 180°) …(i)

But ∠P = ∠R (∵ In a || gm opposite angles are equal) …(ii)

From Eqs. (i) and (ii), we get

∠P = ∠R = 90°

Similarly, ∠Q = ∠S = 90

∴ Each angle of PQRS is 90°.

Hence, PQRS is a rectangle.