Question

In a circle , the 2 chords AB and AC are equal . Prove that the bisector of Angle BAC passes through its centre .
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Please Solve This Problem .....
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Answer 1

hey
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given : AB and AC are two equal chords

to prove : O lies on AD

construction : join BC

proof:

in ∆AMB and ∆AMC

AC=AB
<ABM =<ACM
AM=AM

∆AMB congruent ∆AMC by SAS

<AMB = <AMC = 90°
they both form linear pair and they must be equal by c.p.c.t

hence AM is the perpendicular bisector and even BM=MC( by c.p.c.t )

• by theorem we know that perpendicular bisector of a chord passes through the centre

hence , AM passes through the centre so AD will be the perpendicular bisector of chord BC and BM=MC

AD passes through centre O
O lies on AD

for the diagram refer to the attachment

hope helped
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Answer 2

Step-by-step explanation:

In ΔBAM and CAM,

AB = AC

∠BAM = ∠CAM

AM = AM

ΔBAM ≅ ΔCAM

BM = CM and ∠BMA = ∠CMA

As, ∠BMA + ∠CMA = 180° {Linear pair}

⇒ ∠BMA = ∠CMA = 90°

AM is the perpendicular bisector of chord BC

⇒ AM passes through the center O.

[∵ perpendicular bisector of chord of a circle passes through the centre of circle]

Hence, centre of circle lies on the angle bisector of ∠BAC.

Hope it helps you

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