Question

In the given figure $AB=DC$, $\angle ABC=\angle DCB$. Prove that

(i) $∆ABC\cong ∆DCB$

(ii) $\angle A=\angle D$

(iii) $AC=DB$

Answer 1

STEP 1 : Proving that $∆ABC\cong ∆DCB$

(i)

In $∆ABC$ and $∆DCB$,

$AB=DC$ [ Given ]

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

$BC=CB$ [ Common side ]

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

Hence proved.

STEP 2 : Proving that $\angle A=\angle D$

(ii)

Since, $∆ABC\cong ∆DCB$

As corresponding parts of congruence triangles are equal.

$\therefore \angle A=\angle D$ [ By C.P.C.T.]

Hence proved.

STEP 3 : Proving that $AC=DB$

(iii)

Since, $∆ABC\cong ∆DCB$

As corresponding parts of congruence triangles are equal.

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

Hence proved..