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:
(i) AXCY is a parallelogram
(ii) AX=CY,AY=CX
(iii) $△AYB\cong △CXD$
(iv) $△AXD\cong △CYB$

Answer 1

In $△AXD$ and $△CYB$

$\angle ADX=\angle CBY$ [Alternate interior angle ]

$AD=CB$ [Opposite sides of parallelogram ABCD]

$DX=BY$ [Given]

$\therefore △AXD\cong △CYB$ [Using SAS congruence]

$\therefore AX=CY$ [By CPCT]

Now,

In $△AYB$ and $△CXD$

$\angle ABY=\angle CDX$ [Alternate interior angle ]

$AB=CD$ [Opposite sides of parallelogram ABCD]

$DX=BY$ [Given]

$\therefore △AYB\cong △CXD$ [Using SAS congruence]

$\therefore AY=CX$ [By CPCT]

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

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

Therefore, $AXCY$ is a parallelogram.