# 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 THATANGLE 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!!!

• 69

Given,

∠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.)

In ΔABC,

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

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