Question

In the given figure $AB=DC$ and $\angle ABC=\angle DCB$. Prove that $AC=DB$.

Answer 1

Step 1: Proving that $∆ABC\cong ∆DCB$

In $∆ABC$ and $∆DCB$,

$AB=DC$ [ Given ]

$\angle ABC=\angle DCB$ [Given ]

$BC=CB$ [ Common Side ]

Therefore, by SAS congruence condition $∆ABC\cong ∆DCB$

Step 2: Proving that $AC=DB$

Since corresponding parts of congruent triangles are equal.

$\therefore AC=DB$ [ By C.P.C.T.]

Hence proved.