**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:

Since ABC is an arc of the circle with centre O, which makes an angle AOC = angle AOB + angle BOC

i.e., ∠AOC = ∠AOB + ∠BO = 60° + 30° = 90°

We know ∠AOC = 2 ∠ADC [angle at the centre of a circle is twice the angle at the circumference]

^{ }Therefore ∠ADC = ^{1}/_{2} (∠AOC ) = ^{1}/_{2} x 90° = 45°

**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:

Since OA = OB = AB

Therefore, ∆AOB is an equilateral triangle.

⇒∠AOB = 60°

We know that the angle at the centre of a circle is twice the angle at the circumference.

Therefore, ∠AOB = 2 ∠ACB

⇒∠ACB = ^{1}/_{2} ∠AOB = ^{1}/_{2} x 60° = 30°

Also, we have , ∠ADB = ^{1}/_{2}reflex ∠AOB = ^{1}/_{2}(360° – 60°) = ^{1}/_{2} (300°) = 150°

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 30 °

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

Solution:

Since, angle at the centre of a circle is twice the angle at the circumference.

Therefore, ∠POR = 2 ∠PQR

⇒reflex ∠POR = 2 x 100 ° = 200 °

⇒ ∠POR = 360 ° – 200 ° = 160 °

In ∆OPR, OP = OR [Radii of the same circle]

⇒ ∠OPR = ∠ORP [Angles opposite to equal sides are equal]

and ∠POR = 160 ° [which is proved]

In ∆OPR ,

∠OPR + ∠ORP + ∠PQR = 180 ° [Sum of interior angles of a triangle are equal to 180 °]

⇒160 ° + 2 ∠OPR = 180 °

⇒ 2 ∠OPR = 180 ° – 160 ° = 20 °

⇒ ∠OPR = 10 °

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

Solution:

In ∆ABC,

∠BAC + ∠ABC + ∠BCA = 180 °

⇒ ∠BAC + 69 ° + 31 ° = 180 °

⇒ 180 ° – 100 ° = 80 °

Since the angles in the same segment are equal.

Therefore, ∠BDC = ∠BAC = 80 °

**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:

Given 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°

We have find ∠BAC

Since, ∠CED + ∠CEB = ∠ BED = 180 ° [linear pair]

⇒∠CED + 130 ° = 180 °

⇒∠CED = 180 ° – 130 ° = 50 °

In ∆ECD, we have,

∠EDC + ∠EDC + ∠CED = 180 °

⇒ ∠EDC + 50 ° + 20 ° = 180 °

⇒∠EDC = 180 ° – 50 ° – 20 ° = 110 °

⇒ ∠BDC = ∠EDC = 110 ° [angles in the same segment are equal]

Therefore, ∠BAC = ∠BDC = 110 °

**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:

Given, ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°.

∠CAD = ∠DBC= 70° [Angles in the same segment]

Therefore, ∠DAB = ∠CAD + ∠BAC = 70° + 30° = 100°

But, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]

So, ∠BCD = 180° – 100° = 80°

Now, we have AB = BC

Therefore, ∠BCA = 30° [Opposite angles of an isosceles triangle]

Again, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]

⇒ 100° + ∠BCA + ∠ECD = 180° [∵∠BCD = ∠BCA + ∠ECD]

⇒ 100° + 30° + ∠ECD = 180°

⇒ 130° + ∠ECD = 180°

⇒ ∠ECD = 180° – 130° = 50°

Hence, ∠BCD = 80° and ∠ECD = 50°

**If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.**

Solution:

Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quadrilateral ABCD.

We have to prove quadrilateral ABCD is a rectangle.

We know, all the radius of the same circle are equal.

Therefore, OA = OB = OC = OD

⇒OA = OC = ^{1}/_{2} AC

⇒OB = OC = ^{1}/_{2} BD

⇒AC = BD

Thus, the diagonals of the quadrilateral ABCD are equal and bisect each other.

Quadrilateral ABCD is rectangle.

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

Solution:

Given : A trapezium ABCD in which AB || CD and AD = BC.

To Prove : ABCD is a cyclic trapezium.

Construction : Draw DE ⊥ AB and CF ⊥ AB.

Proof : In ∆DEA and ∆CFB, we have

AD = BC [Given]

∠DEA = ∠CFB = 90° [DE ⊥ AB and CF ⊥ AB]

DE = CF [Distance between parallel lines remains constant]

∴ ∆DEA ≅ ∆CFB [RHS axiom]

⇒ ∠A = ∠B …(i) [CPCT]

and, ∠ADE = ∠BCF ..(ii) [CPCT]

Since, ∠ADE = ∠BCF [From (ii)]

⇒ ∠ADE + 90° = ∠BCF + 90°

⇒ ∠ADE + ∠CDE = ∠BCF + ∠DCF

⇒ ∠D = ∠C ……………………………….(iii)

[∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C]

∴ ∠A = ∠B and ∠C = ∠D [From (i) and (iii)] (iv)

∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°]

⇒ 2(∠B + ∠D) = 360° [Using (iv)]

⇒ ∠B + ∠D = 180°

⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.

⇒ ABCD is a cyclic trapezium

**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:

Since angles in the same segment are equal.

Therefore, ∠ACP = ∠ABP ……………….(1)

and ∠QCD = ∠QBD ……………………(2)

Also ∠ABP = ∠QBD ……………………..(3) [vertically opposite angles]

Therefore, from (1), (2) and (3), we have

∠ACP = ∠QCD

**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 AB and AC of ∆ABC as diameters . The circle intersect at D

To prove: D lies on BC

Construction:

Join A and D

Proof: since AB and AC are the diameters of the two circles[given]

Therefore, ∠ADB = 90° [angles are in a semi circle]

and ∠ADC = 90° [angles in a semi circle]

On adding, we get,

⇒∠ADB + ∠ADC = 90 + 90 = 180

⇒BDC is a straight line

Hence D lies on BC

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

Solution:

∆ABC and ∆ADC are right angles with common hypotenuse AC. Draw a circle AC as diameter passing through B and D . Join BD

Clearly, ∠CAD = ∠CBD [since angles in the same segment are equal]

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

Solution:

Given: ABCD is a parallelogram inscribed in a circle

To prove: ABCD is a rectangle

Proof: Since ABCD is a cyclic parallelogram

Therefore, ∠A + ∠C = 180° ………………..(1)

But ∠A = ∠C ………………..(2)

From (1) and (2), we have

∠A = ∠C = 90°

Similarly, ∠B = ∠D = 90°

Therefore, each angle of ABCD is of 90°

Hence, ABCD is a rectangle.

## One response to “Circles – Exercise 10.5 – Class IX”

[…] Circles – Exercise 10.5 […]

LikeLiked by 1 person