Question

explain theorem 10.10,10.9,10.8 class 9 chapter circles​

Answer 1

Answer:

Theorem 10.8 :

The angle subtended by an arc at the centre is double the angle

subtended by it at any point on the remaining part of the circle.

Proof : Given an arc PQ of a circle subtending angles POQ at the centre O and

PAQ at a point A on the remaining part of the circle. We need to prove that

∠ POQ = 2 ∠ PAQ.

Fig. 10.28

Consider the three different cases as given in Fig. 10.28. In (i), arc PQ is minor; in (ii),

arc PQ is a semicircle and in (iii), arc PQ is major.

Let us begin by joining AO and extending it to a point B.

In all the cases,

∠ BOQ = ∠ OAQ + ∠ AQO

because an exterior angle of a triangle is equal to the sum of the two interior opposite

angles.

Theorem 10.9 :

Angles in the same segment of a circle are equal.

Again let us discuss the case (ii) of Theorem 10.8 separately. Here ∠PAQ is an angle

in the segment, which is a semicircle. Also, ∠ PAQ =

1

2 ∠ POQ =

1

2 × 180° = 90°.

If you take any other point C on the semicircle, again you get that

∠ PCQ = 90°

Therefore, you find another property of the circle as:

Angle in a semicircle is a right angle

Theorem 10.10 :

If a line segment joining two points subtends equal angles at

two other points lying on the same side of the line containing the line segment,

the four points lie on a circle (i.e. they are concyclic).

You can see the truth of this result as follows:

In Fig. 10.30, AB is a line segment, which subtends equal angles at two points C and

D. That is

∠ ACB = ∠ ADB

To show that the points A, B, C and D lie on a circle

let us draw a circle through the points A, C and B.

Suppose it does not pass through the point D. Then it

will intersect AD (or extended AD) at a point, say E

(or E′).

If points A, C, E and B lie on a circle,

∠ ACB = ∠ AEB (Why?)

But it is given that ∠ ACB = ∠ ADB.

Therefore, ∠ AEB = ∠ ADB.

This is not possible unless E coincides with D. (Why?)

Similarly, E′ should also coincide with D.