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.
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]
∠AQC we can break it in two parts ∠AQC=∠AQB+∠BQC
[∠AQB=∠ACB angles subtend by same chord AB,similarly ∠BQC=∠BAC angles subtend by same chord BC ]
from (1) and (2),
which is the required result.
hope this helps you.
∠ABQ = ∠ACQ
∴ A, B, C and Q are concyclic points.
(If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, then the four points are con cyclic.)
AB = AC
∴ ∠ACB = ∠ABC ....(1) (Equal sides have equal angles opposite to them)
∠ACB + ∠ABC + ∠BAC = 180° (Angle sum property)
∴ 2∠ACB + ∠BAC = 180° (Using (1))
⇒ 2∠ACB = 180° – ∠BAC
⇒ ∠ACB = 90 – ∠BAC ...(2)
∠ACB = ∠AQB ...(3) (Angles in the same segment are equal)
∠BAC = ∠BQC ...(4) (Angles in the same segment are equal)
Adding (3) and (4), we have
∠BAC + ∠ACB = ∠AQB +∠BQC
∴ ∠BAC + ∠ACB = ∠AQC
∠ACB =∠ AQC – ∠BAC ...(5)
From (2) and (5), we have
∠AQC – ∠BAC = 90 ° – ∠BAC
∴ ∠AQC = 90° + ∠BAC – ∠BAC
⇒ ∠AQC = 90° + ∠BAC