Question

In the given figure, AP is bisector of $\angle A$ and CQ is bisector of $\angle C$ of parallelogram ABCD.
Prove that APCQ is a parallelogram

Answer 1

Consider $△ADP$ & $△QBC$

$\angle D=\angle B$ (Opposite side of parallelogram.)

$\therefore DA=BC$ (Opposite side of parallelogram.)

$\angle x=\angle x$ [bisector of opposite angle.]

$\therefore$ By $ASA$ congruence rule. $[△ADP\cong △QBP]$

then by $CPCT,$ $DP=QB⟶\left(1\right)$

$\angle x=\angle x$ [bisector of opposite angle.]

$△ADP\cong △QBP$ [by $ASA$]

$AB=DC$ [opposite side of a parallelogram.]

$AQ+QB=DP+PC$ [$QB=DP$ from $1$]

$AQ=PC⟶\left(2\right)$

Hence distance between $AP$ and $QC$ are equal at all place hence proved $AP\parallel QC⟶\left(3\right)$

$\therefore AQCP$ is parallelogram.