Class 9 Maths

# Circle

## Exercise 10.6

### Part 2

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) So, AB = CD

Or, arc AB = arc D ………………(1)

Similarly, arc AD = arc BC ………(2)

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 Proof: Since angles in the same segment are equal

So, ∠FDA = ∠FCA ………………..(1)

∠EDA = ∠EBA ……………………..(2)

∠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 ∠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 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

So, ∠PAC = 1/2 θ