# Circle

## Exercise 10.6

### Part 2

Question 4: Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

**Answer:** This figure shows ∠ABC and its sides are extended respectively to D and E so that AD and CE are equal chords of the circle.

**To Prove:** ∠ABC = `1/2` (∠DOE - ∠AOC)

Let us join A to E

**Proof:** In ΔBAE

∠DAE = ∠ABC = ∠AEC ………….(1)

(External angle of triangle is equal to sum of angles of opposite vertices)

Chord DE subtends ∠DOE at centre and ∠DAE in remaining part of the circle.

So, ∠DAE = `1/2` ∠DOE ……..(2)

Similarly, ∠AEC = `1/2` ∠AOC ………….(3)

From equations (1), (2) and (3)

`1/2∠DO\E=∠AB\C+1/2∠AO\C`

Or, `∠AB\C=1/2∠DO\E-1/2∠AO\C`

Or, `∠AB\C=1/2(∠DO\E-∠AO\C)` Proved

Question 5: Prove that the circle drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.

**Answer:** This figure shows a rhombus ABCD in which diagonals AC and BD intersect at O.

Let us draw a circle by taking AB as diameter. Let us draw PQ||DA and RS||AB. Both PQ and RS are passing through O. Now, P, Q, R and S are midpoints of DC, AB, AD and BC respectively.

Since Q is the midpoint o AB

Hence, AQ = QB ………(1)

Since AD = BC (sides of rhombus are equal)

So, `1/2AD=1/2BC`

Or, RA = OQ ………(2)

Since PQ is parallel to AD and AD = BC

So, AB = AD (sides of rhombus)

Or, `1/2AB=1/2AD`

Or, AQ = AR ………….(2)

From equations (1), (2) and (3)

AQ = QB = OQ

This means that a circle drawn with Q as centre will pass through A, B and O

So, it is proved that the circle passes through the intersection O of the diagonals from rhombus ABCD.

Question 6: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD

**Answer:** This figure shows a parallelogram ABCD and a circle passes through A,B and C but intersects CD at point E.

Given, ABCE is a cyclic quadrilateral

So, ∠AEC + ∠B = 180° …………………..(1)

ABCD is a parallelogram

So, ∠D = ∠B ……………..(2) (Opposite angles of parallelogram are equal)

From equations (1) and (2)

∠AEC + ∠D = 180° ……………………….(3)

But ∠AEC + ∠AED = 180° (Linear pair of angles) …………………… (4)

From equations (3) and (4)

∠D = ∠AED

This means that base angles of ΔADE are equal.

So, AE = AD (Sides opposite equal angles)