Question

In the given figure, if DC = AB and
AD = BC, prove that Angle ABD is congulruent to angle CDB and
Angle ABD = Angle CDB.​

Answer 1

Step-by-step explanation:

angle ABD = angle CBD

DB is common

AB=DC

so the 2 ∆ are congruent by SAS rule

Hope this helps u

Answer 2

$\huge\sf\blue{Given}$

✭ DC = AB

✭ AD = BC

$\rule{110}1$

$\huge\sf\gray{To\;Prove}$

☆ $\sf\triangle ABD \cong \triangle CDB$

☆ $\sf\angle ABD = \angle CDB$

$\rule{110}1$

$\huge\sf\purple{Steps}$

So with the given info,

➝ $\sf AB = DC \qquad\bigg\lgroup Given\bigg\rgroup$

➝ $\sf AD = BC \qquad\bigg\lgroup Given \bigg\rgroup$

➝ $\sf DB = DB \qquad\bigg\lgroup Given \bigg\rgroup$

$\sf\therefore \triangle ABD \cong \triangle CDB$

So now,

$\sf\dashrightarrow \angle ABD = \angle CDB \qquad\bigg\lgroup CPCT \bigg\rgroup$

Hence Proved!!

$\rule{170}3$