Question

In the given figure, $ABCD$ is a parallelogram. And $X$ and $Y$ are points on the diagonal $BD$ such that $DX=BY$. Prove that:
$\begin{array}{c}\text{(i) AXCY is a parallelogram}\\ \text{(ii)}AX=CY,AY=CX\\ \text{(iii)}△AYB\cong △CXD\\ \text{(iv)}△AXD\cong △CYB\end{array}$

Answer 1

In $△AYB$ and $△CXD$

$AB=CD$ (opposite side of the parallelogram)

$\angle ABY=\angle CDX$ (property of parallel lines)

$DX=BY$ (given)

so $△AYB\cong △CXD$

so $AY=CX$ ....(1)

similarly in $△AXD$ and $△CYB$

$AD=BC$ (opposite side of the parallelogram)

$\angle ADX=\angle CBY$ (property of parallel lines)

$DX=BY$ (given)

so $△AXD\cong △CYB$

so $AX=CY$ ....(2)

from (1),(2) $AXCY$ is a parallelogram