ABC IS A TRIANGLE IN WHICH AB=AC. P IS A POINT ON AC. THROUGH C A LINE IS DRAWN TO INTERSECT BP PRODUCED AT Q SUCH TAHT ANGLE ABQ=ANGLE ACQ. PROVE THAT

ANGLE AOC=90 DEGREES+ HALF OF ANGLE BAC.

given: ABC is a triangle, where AB=AC. P is any point on AC.

∠ABQ=∠ACQ

TPT: ∠AQC=90+1/2∠BAC

proof:

since ∠ABQ=∠ACQ

since line segment AQ subtends equal angles ∠ABQ and ∠ACQ at points B and C respectively on the same side, therefore A, B, C and Q are concylic points.

∠ABC=∠ACB [angles opposite to equal sides are equal]

∠BAC=180-(∠ABC+∠ACB)

∠BAC=180-2∠ACB

2∠ACB=180-∠BAC

∠ACB=90-1/2∠BAC...............(1)

∠AQC we can break it in two parts ∠AQC=∠AQB+∠BQC

∠AQC=∠ACB+∠BAC .......(2)

[∠AQB=∠ACB angles subtend by same chord AB,similarly ∠BQC=∠BAC angles subtend by same chord BC ]

from (1) and (2),

∠AQC=90-1/2∠BAC+∠BAC

∠AQC=90+1/2∠BAC

which is the required result.

hope this helps you.

cheers!!!

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