Question

14. AD is the bisector of √A of a triangle ABC.
P is any point on AD. Prove that the
perpendiculars drawn from P on AB and AC
are-equal.​

Answer 1

Answer:

Triangles are similar...

<CAD = <DAB {AD bisects <CAB}

Perpendiculars from P. {Say: PE & PF}

<EPA = <FPA {90 degrees - <CAB/2}

AP is common.

Hence triangles are congruent &

PE = PF

Step-by-step explanation: