Question

in the given figure AB and DC are both perpendicular to line segment AD. BC intersects AD at P and P s midpoint of AD. prove that 1) AB=CD 2) P is midpoint of BC​

Answer 1

Step-by-step explanation:

clearly from the figure

angle PAB = angle PDC (as both are 90 degrees)

PA = PB (Given)

angle APB = Angle DPC ( vertically opposite angles)

therefore

APB is congruent to DPC by ASA

therefore

1. AB = DC ( cpct)
2. PB = PC so p is mid point

Answer 2

Step-by-step explanation:

Given Data:

$<PAB=<PDC \\ \: right \: angle \: triangle$

$AP=PD \: \\ cb \: side \: cut \: ad \: side \: at \: mid$

also

$<BPA=<DPC \: \\ by \: vertical \: angles \: rule$

as

by

A.S.A=A.S.A

rules the two triangles are congruent

so

Answer:

$AB=CD$

also

$P \: is \: the \: mid \: point \: of \: BC$