Question

In quadrilateral $ABCD$ of the given figure $X$ and $Y$ are point on diagonal $AC$ such that $AX=CY$ & $BXDY$ is a parallelogram. Show that $ABCD$ is a parallelogram.

Answer 1

$BXDY$ is a parallelogram, $BD$ and $XY$ are diagonals of parallelogram.

We know that diagonals of parallelogram bisect each other.

$XO=YO\dots \left(i\right)$

$DO=BO\dots \left(ii\right)$

$AX=CY\dots \left(iii\right)$

Adding $\left(i\right)$ and $\left(iii\right)$, we have

$XO+AX=YO+CY$

$\Rightarrow$ $AO=CO\dots \left(iv\right)$

From $\left(ii\right)$ and $\left(iv\right)$, we have

$AO=CO$ and $DO=BO$

This implies that $AC$ and $BD$ bisect each other but $AC$ and $BD$ are diagonals of quadrilateral $ABCD$ hence, $ABCD$ is a parallelogram.