Theorem 10.8 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 10.8 class 9 Maths which is related to Chapter 10 Circles Class 9 Mathematics. After understanding theorem 10.8, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 

 Theorem 10.8 Class 9

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.

Given 

An arc PQ of a circle subtending angles POQ at the centre O and PAQ at point A on the remaining part of the circle.

To Prove

∠POQ=2∠PAQ 

Proof 

Consider the three different cases as given in figure (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 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.

Also, in ∆OAQ,     OA=OQ  (Radii of a circle)

Therefore,  ∠OAQ=∠OQA          (theorem 7.5)

This gives ∠BOQ=2∠OAQ              ………(1)

Similarly,  ∠BOP=2∠OAP                ………(2)

From (i) and (ii), ∠BOP+∠BOQ=2(∠OAP+ ∠OAQ)

This is the same as ∠POQ=2∠PAQ     ………(3)

For case (iii), where PQ is the major arc (3) is replaced by reflex angle ∠POQ=2∠PAQ 

 Related Topics

1. Theorem 10.9

2. Theorem 10.7