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!!!