Answer the question attached
MATHS, CIRCLES

Dear student,
Please find below the solution to the asked query :

GIVEN: ABC is an equilateral triangle inscribed in a circle having the centre at O.

P be any point on the minor arc BC which does not coincide wit B or C.

TO PROVE : PA is the angle bisector of ∠BPC.

CONSTRUCTION : Join AP, BP and CP and Join OA, OB and OC.

PROOF :

Since, ABC is an equilateral , thenAB = BC = AC  .....1Since, chord AB = chord AC  using 1AOB = AOC  Equal chords subtend equal angles at the centre    .......2We know that, angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.Now, consider the arc AB, thenAOB = 2APB        ..........3Consider the arc AC, thenAOC = 2APC        .........4Substituting the values of AOB and AOC in 2, we get    2APB = 2APCAPB = APCAP is the angle bisector of BPC.

Hope this information will clear your doubt about this topic.                             
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