Question

ABC is a triangle in which AB = AC. P is any point on BC.
(i) Prove that AB>AP. (ii) If angle BAP = 54° and angle PAC = 24°, find angle APB.

Answer 1

Answer:

We have ABCD a cyclic quadrilateral and BC II AD

The sum of opposite angles of cyclic quadrilateral is 180 degree

This gives angle ABC=180-110=70 degree

The triangle ABC has angle ABC=70  and angle BAC=50

           Hence angle BCA=180 - 70 - 50 = 60 degree

 We know BC II AD and angles BCA and DAC are alternate interior angles

 Hence  angle BCA = DAC=60 degree

  Angle DAC=60 degree

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