Question

In figure ABCD is a square p and q are the midpoints of sides AB and BC respectively prove that PB is equal to QD

Answer 1

$\orange{\underline{We \:know \:that \:ABCD\: is\: a \:square }}$

Also note that P and Q are the mid points of DA & BC

AD = BC

DP = PA ......( P is the mid point )

CQ = QB ....( Q is the mid point )

DA = 2 AP

CB = 2 CQ

$\cancel{2}\times AP = \cancel{2}\times CQ \\\implies AP=CQ$ .....(1)

$\red{\underline{Now\: in \: \triangle \:PAB\: and \: \triangle\: QCD }}$

$AP\:=\:CQ \implies proved \\\\ \angle PAB = \angle QCD \implies 90\degree\\\\ AB = CD \implies side\: of \:square$

$\triangle PAB = \triangle QCD$

$\boxed{\pink{By\: SAS \:Property}}$

By CPCT

$\boxed{\pink{DQ = BP}}$

Hence Proved