**Circles**

**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:**We have a circle with centre O, such that

∠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 = 12 (∠AOC) = 12(90°) = 45°

**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:**

We have a circle having a chord AB equal to radius of the circle.

∴ AO = BO = AB

⇒ ∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle is 60°.

⇒ ∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.

Similarly, ∠ADB = 12 [∠AOB] = 12 x 60° = 30°

Hence, the angle subtended by the chord on the major arc = 30°

**Question 3.In figure, ∠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 pn 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]

⇒ ∠OPR = 20∘2 = 10°

**Question 4.In figure, ∠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°

**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:**

∠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°

**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°

lso, ∠DBC = 70° [Given]

In ∆BCD, we have

∠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°

**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:**

Since AC and BD are diameters.

⇒ AC = BD …(i) [All diameters of a circle are equal]

Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]

Similarly, ∠ABC = 90°, ∠BCD = 90°

and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have

AC = BD [From (i)]

AB = BA [Common hypotenuse]

∠ABC = ∠BAD [Each equal to 90°]

∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]

⇒ BC = AD [C.P.C.T.]

Similarly, AB = DC

Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.

∴ ABCD is a rectangle.

**Question 8.If the non – parallel sides of a trapezium are equal, prove that it is cyclic.**

**Solution:**

We have a trapezium ABCD such that AB ॥ CD and AD = BC.

Let us 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), we have 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.

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

∴ ∠ACP = ∠ABP …(i)

Similarly, ∠QCD = ∠QBD …(ii)

Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]

∴ From (i), (ii) and (iii), we have

∠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:**

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

Let us 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.

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

**Solution:**

We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC

∴ Both the triangles are in semi-circle.

Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.

Join BD.

DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

**Question 12.Prove that a cyclic parallelogram is a rectangle.**

**Solution:**

We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.

∴ Sum of its opposite angles is 180°.

⇒ ∠A + ∠C = 180° …(i)

But ∠A = ∠C …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

⇒ Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.