Question

In the figure AB and AC are two equal chords of a circle. Prove that the bisector of angle BAC passes through the centre of the circle...
Plz write the answer and don't write useless things... ​

Answer 1

Answer: Proved

Step-by-step explanation:

Let the two chords AB and AC are chords

centre of the circle O lies on the bisector of the ∠BAC

Now join BC

let the bisector of ∠ BAC intersect BC at P

Now from ΔAPB and ΔAPC

AB =AC

BAP =CAP

AP=AP

ΔAPB ≅  Δ APC

BP=CP and ∠APB=∠ APC     ( By CPCT)

Now ∠APB + ∠APC = 180° ( linear pair)

2 ∠APB = 180

∠APB = 180 / 2

∠APB = 90°

So. BP =CP  ,

Hence AP is the perpendicular bisector of the chord BC and AP passes through the centre of the circle .

Answer 2

Step-by-step explanation:

• Show that ΔABM is congruent to ΔAMC.

• By C.P.C.T prove that angle ABM is equal to angle AMC & BM=MC.

• Then prove that AD is the perpendicular bisector of Chord BC.
• Theorem : only the line drawn through the centre of a circle to bisect the chord is perpendicular to the chord.