Question

Two chords AB and AC lie on the opposite sides of the centre O of a circle. If angle ACO = 15 and angle ABO = 45, then find the difference between the measures of BAC and BOC.​

Answer 1

Answer:

Step-by-step explanation:

Given : AB and AC are two equal chords of C (O, r).

Required to prove : Centre, O lies on the bisector of ∠BAC.

Construction : Join BC, Let the bisector of ∠BAC intersects BC in P.

Proof :

In Δ and ΔAPC,

AB=AC [Given]

∠BAP=∠CAP [Given]

AP=AP [Common]

Hence, ΔAPB≅ΔAPC by SAA congruence criterion

SO, by CPCT we have

BP=CP and ∠APB=∠APC

And

∠APB+∠APC=180  

∘

 [Linear pair]

2∠APB=180  

∘

[∠APB=∠APC]

∠APB=90  

∘

 

Now, BP=CP and ∠APB=90  

∘

 

Therefore, AP is the perpendicular bisector of chord BC.

Hence , AP passes through the centre, O of the circle.