Circle

Exercise 10.6

Part 2

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)

Quadrilateral in Circle

From equations (3) and (4)

∠D = ∠AED

This means that base angles of ΔADE are equal.

So, AE = AD (Sides opposite equal angles)



Question 7: AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Answer: This figure shows a circle in which two chords AC and BD bisect each other and point of bisection is O

To Prove: (i) AC and BD are diameters and(ii) ABCD is a rectangle

Construction: Join AB, BC, CD and DA

Proof: (i) In ΔAOB and ΔCOD

AO = CO (O is the midpoint of AC)

BO = DO (O is the midpoint of BD)

∠AOB = ∠COD (vertically opposite angles)

So, ΔAOB ≅ΔCOD (SAS theorem)

Quadrilateral in Circle

So, AB = CD

Or, arc AB = arc D ………………(1)

Similarly, arc AD = arc BC ………(2)

Adding these equations, we get

Arc AB + arc AD = arc CD + arc BC

So, BD divides the circle into two equal parts

Hence, BD is a diameter

Similarly, AC is a diameter

Now, let us solve question (ii).

We have proved ΔAOB ≅ δCOD

Or, ∠OAB = ∠OCD

Or, ∠CAB = ∠ACD

Or, AB||DC Similarly, AD||BC can be proved

This proves that ABCD is a parallelogram

∠DAB = ∠DCB (Opposite angles of parallelogram are equal)

But ∠DAB + ∠DCB = 180° (Opposite angles of cyclic quadrilateral are complementary)

So, ∠DAB = 90° = ∠DCB

So, it is proved that ABCD is a rectangle

Question 8: Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are `90°-1/2A`, `90°-1/2B` and `90°-1/2C`

Answer: This figure shows a triangle ABC inscribed in a circle and bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.

Construction: Join DE, EF and FD

Star in Circle

Proof: Since angles in the same segment are equal

So, ∠FDA = ∠FCA ………………..(1)

∠EDA = ∠EBA ……………………..(2)

Adding these equations, we get

∠FDA + ∠EDA = ∠FCA + ∠EBA

`=1/2∠C+1/2∠B`

`=1/2(∠C+∠B)=1/2(180°-∠A)`

`=90°-(∠A)/2`

Similarly, ∠FED = `(90°-(∠B)/2)`

And, ∠EFD = `(90°-(∠C)/2)`

So, angles of ΔDEF are as follows:

`(90°-(∠A)/2)`, `(90°-(∠B)/2)`, and `(90°-(∠C)/2)`


Question 9: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Answer: This figure shows two congruent circles intersecting each other at points A and B. A line segment passing through A, meets the circles at points P and Q. Let us draw the common chord AB

Two Congruent Circles

∠APB = ∠AQB (angles subtended by equal chords in congruent circles)

In ΔPBQ

∠AQB = ∠APB

So, BP = BQ proved

Question 10: In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer: This figure shows ΔABC and O is the centre of its circumcircle. Perpendicular bisector of BC passes through O and intersects the circle at P. Now, join OB and OC

We need to prove that AP is the bisector of ∠BAC

Triangle in Circle

Let us assume that arc BC makes angle θ at circumference

Then ∠BOC = 2θ

(Because angle at centre is double the angle made by an arc at circumference)

Also, in ΔBOC

OB = OC (Radii)

OP is perpendicular bisector of BC

So, ∠BOP = ∠COP = θ

Arc CP makes angle θ at O

So, ∠PAC = `1/2` θ

(Because angle at centre is double the angle made by an arc at circumference)

This means that AP is the bisector of ∠BAC



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