Question

In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that ΔAFD≅ΔBEC.

Answer 1

seg AD=seg BC
angle FAD=angle CBE.
seg AF =seg BE. ............... property of parallel lines
By SAS test
∆ AFD=∆BEC

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  Given that in the above figure, ABCD is a parallelogram ABEF is a rectangle.

We are to show that ΔAFD≅ΔBEC.

We know that

the measures of the opposite sides of a parallelogram and a rectangle are equal.

So, we have

$AB=EF~~\textup{and}~~AB=CD.$

That is,

$EF=CD\\\\\Rightarrow EF-DE=CD-DE~~~~~~~~~~~[\textup{subtracting the length of DE from both sides}]\\\\\Rightarrow DF=EC.$

Also, $AF=BE,\\\\AD=BC.$

So, in triangles AFD and BEC, we have

$AF=BE,\\\\AD=BC,\\\\DF=EC.$

Therefore, by SSS(side-side-side postulate), we get

ΔAFD≅ΔBEC.

Hence showed.