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.
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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:
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