## Circles

**Chapter 10**

**Exercise 10.5**

**EX 10.5 QUESTION 1.**

**In Fig. 10.36, A,B and C are three points on a circle with centre O such that BOC = 30° and AOB = 60°. If D is a point on the circle other than the arc ABC, find ADC.**

**Solution:**

∠AOB = 60° and ∠BOC = 30°

∵∠AOB + ∠BOC = ∠AOC

∴ ∠AOC = 60° + 30° = 90° [The angle subtended by an arc at the circle is half the angle subtended by it at the centre.]

∴ ∠ADC =(½)∠AOC

**EX 10.5 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:**

In ∆OAB

OA = AB [Given]

OA = OB [radii of the circle]

∴ OA = OB = AB

∴ ∠OAB = 60° [Each angle of an equilateral triangle is 60°]

⇒ ∠AOB = 2∠ADB [ The angle subtended by an arc at the centre is double the angle subtended by it at any ]

⇒ ∠ADB = ½∠AOB ⇒ ∠ADB = ½ x 60°= 30°

ABCD is a cyclic quadrilateral

∠ACB + ∠ADB = 180° [the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]

∠ACB + 30° = 180°

⇒ ∠ACB 180° – 30° = 150°

**EX 10.5 QUESTION 3.**

**In Fig. 10.37, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.**

**Solution:**

[The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.]

∴ Reflex ∠POR = 2∠PQR

But ∠PQR = 100°

∴ reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

⇒ ∠POR = 360° – 200°

⇒ ∠POR = 160°

Since, OP = OR [Radii of the same circle]

∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal]

Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle is 180°]

⇒ ∠OPR + ∠ORP + 160° = 180°

⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]

⇒ 2∠OPR = 20° = ∠OPR = 10°

**EX 10.5 QUESTION 4.**

** In Fig. 10.38, ABC = 69°, ACB = 31°, find BDC.**

**Solution:**

In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 69° + 31° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100° = 80°

[Since, angles in the same segment are equal.]

∴∠BDC = ∠BAC ⇒ ∠BDC = 80°

**EX 10.5 QUESTION 5.**

**In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find BAC.**

**Solution:**

∠BEC = ∠EDC + ∠ECD [Sum of interior opposite angles is equal to exterior angle]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20° = 110°

⇒ ∠BDC = 110°

Since, angles in the same segment are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BAC = 110°

**EX 10.5 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:**

Since angles in the same segment of a circle are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BDC = 30°

∠DBC = 70° [Given]

In ∆BCD,

∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]

⇒ ∠BCD + 70° + 30° = 180°

⇒ ∠BCD = 180° -100° = 80°

Now, in ∆ABC,

AB = BC [Given]

∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]

⇒ ∠BCA = 30° [∵ ∠B AC = 30°]

Now, ∠BCA + ∠BCD = ∠BCD

⇒ 30° + ∠ECD = 80°

⇒ ∠BCD = 80° – 30° = 50°

**EX 10.5 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:**

AC is the diameter of the circle.

⇒ ∠ADC =90° and ∠ABC = 90° …(i) [angle in semicircle is at 90°]

BD is the diameter of the circle.

Similarly, ∠BAD = 90° and ∠BCD = 90° …(ii) [angle in semicircle is at 90°]

and ∠CDA = 90°

from equation (i and (ii)

∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°

∴ ABCD is a rectangle.

**EX 10.5 QUESTION 8.**

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

**Solution:**

In trapezium ABCD such that AB ॥ CD and AD = BC.

Draw BE ॥ AD such that ABED is a parallelogram.

∵ The opposite angles and opposite sides of a parallelogram are equal.

∴ ∠BAD = ∠BED …(i)

and AD = BE …(ii)

But AD = BC [Given] …(iii)

∴ From (ii) and (iii), BE = BC

⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]

Now, ∠BED + ∠BEC = 180° [Linear pair]

⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]

i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.

∴ ABCD is cyclic.

⇒ The trapezium ABCD is cyclic.

**EX 10.5 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 Fig. 10.40). Prove that ∠ ACP = ∠ QCD.**

**Solution:**

∠ACP = ∠ABP …(i) [angles in the same segment of a circle are equal.]

∠ABP = ∠QBD …(ii) [Vertically opposite angles]

∠QCD = ∠QBD …(iii)

∴ From (i), (ii) and (iii), we have

∠ACP = ∠QCD

**EX 10.5 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:**

In ∆ABC, and two circles described with the diameter as AB and AC respectively. They intersect at a point D, other than A.

Join A and D.

∵ AB is a diameter.

∴∠ADB is an angle formed in a semicircle.

⇒ ∠ADB = 90° ……(i)

Similarly, ∠ADC = 90° ….(ii)

Adding (i) and (ii), we have

∠ADB + ∠ADC = 90° + 90° = 180°

i. e., B, D and C are collinear points.

⇒ BC is a straight line. Thus, D lies on BC.

**EX 10.5 QUESTION 11.**

**ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠CBD.**

**Solution:**

Given: Triangle ABC and ADC are two triangles on a common base AC.

To prove : ∠CAD = ∠CBD

Proof: ∆ABC and ∆ADC are on common base BC and ∠BAC = ∠BDC.

Points A, B, C and D lie on the same circle.

[If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]

∴ ∠CAD = ∠CBD

**EX 10.5 QUESTION 12.**

**Prove that a cyclic parallelogram is a rectangle.**

**Solution:**

Given: ABCD is a cyclic quadrilateral.

⇒ ∠A + ∠C = 180° …(i) [∴ Sum of its opposite angles is 180°.]

But ∠A = ∠C …(ii) [Opposite angles of a parallelogram are equal]

From (i) and (ii),

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

⇒ Each angle of the parallelogram ABCD is 90°.

∴ ABCD is a rectangle.