Question

In the given figure, AB and AC are chords of a circle, if ACO = 15° and ABO = 45°, then find the difference between the measures of BAC and BOC.​

Answer 1

Answer:

In the given figure, AB and AC are chords of a circle, if ACO = 15° and ABO = 45°, then find the difference between the measures of BAC and BOC.

Answer 2

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 ZBAC intersects BC in P.

Proof:

In A and AAPC,

AB=AC [Given]

ZBAP=<CAP [Given]

AP=AP [Common]

Hence, AAPB=AAPC by SAA congruence criterion

SO, by CPCT we have

BP=CP and <APB=ZAPC

And

ZAPB+ZAPC=180

[Linear pair]

22APB=180

о

[ZAPB=ZAPC]

ZAPB=90

Now, BP-CP and ZAPB-90

Therefore, AP is the perpendicular bisector of chord BC.

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