Question

Prove the Theorem 10.8 Circles Class 9th NCERT?

Answer 1

Theorem 10.8  class IX

"The angle subtended by arc at the centre is double than angle subtended by it at any point on the remaining part of the circle."

Proof:
Given arc PQ of circle subtending angles POQ at the centre O
PAQ at a point A on the remaining part of the circle

RTP: ∠POQ=2∠PAQ

Solution:
let us consider three different cases as the figures attached.
Case 1:Arc PQ is minor
Case 2Arc PQ is semicircle
case 3:Arc PQ is major

Let us join AO and extending it to point BIn all the 3 cases we have,∠BOQ=∠OAQ+∠AQO
because exterior angle of a traingle is equal to sum of two interior opposite angles

Also in triangle OAQ
OA=OQ (radii of circle)

Therefore ∠ OAQ=∠OQA [ ACCORDING TO THEOREM 7.5]
⇒∠BOQ=2∠OAQ---------1

similarly ∠BOP=2∠OAP--------------2

from 1 and 2 we get

∠BOP+∠BOQ=2∠OAP+∠OAQ]

∠POQ=2∠PAQ

In case 3: arc Pq is major:
equation 3 is replaced by reflex angle ∠ POQ=2∠PAQ

Answer 2

Answer:

mathe class 9 theorem 10.8