Question

ABCD is a stright line with AB=BC=CD. P is a point not on the straight line such that PB=PC . Prove PA=PD

Answer 1

Draw a line ABCD

Such that       AB = BC= CD

Taking Point P and joining with Points B and C

it is given that     PB = PC

Now in Triangle ABP and DPC

AB = CD
PB = PC
and
Angle PBA = Angle PCD
Thus by SAS

Triangle are similar

PA = PD

Hence Proved

Answer 2

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