Question

In fig. ABCD is a parallelogram and X and Y are points on the diagonal BD such that DX=BY. Prove that XAYC is a parallelogram.

Answer 1

In $△AXD$ and $△CYB$

$\angle ADX=\angle CBY$ (alternate interior angles for $BC\parallel AD$)

$AD=CB$(opposite sides of parallelogram $ABCD$)

$DX=BY$(given)

$\Rightarrow △AXD\cong △CYB$ by SAS congruency rule.

$\Rightarrow AX=CY$ by C.P.C.T

Now, in $△AYB$ and $△CXD$,

$\angle ABY=\angle CDX$ (alternate interior angles for $AB\parallel CD$)

$AB=CD$(opposite sides of parallelogram $ABCD$)

$DX=BY$(Given)

$\Rightarrow △AYB\cong △CXD$ by SAS congruency rule.

$\Rightarrow AY=CX$by C.P.C.T

From the result we obtained $AX=CY$ and $AY=CX$

Since opposite sides of quadrilateral $AXCY$ are equal to each other.

$\therefore XACY$ is a parallelogram.

Hence proved