Question

Question 2
In figure, D and E are points on side BC of the a $\Delta ABC$ such that BD = CE and AD = AE, show that $\Delta ABD\cong \Delta AE$ show that $\Delta ABD\cong \Delta ACE$

Answer 1

Given D and E are the points on side BC of a such that BD = CE and AD = AE to show $\Delta ABD\cong \Delta ACE$
proof we have AD =AE [given]
$\Rightarrow \angle ADE=\angle AED....\left(i\right)$
[since, angles opposite to equal sides are equal]
we have $\angle ADB+\angle ADE={180}^{\circ }$ [linear pair axiom]
$\Rightarrow \angle ADB={180}^{\circ }-\angle ADE$
$={180}^{\circ }-\angle AED$ [from Eq.(i)]
in $\Delta ABD$ and $\Delta ACE$
$\angle ADB=\angle ACE[\because \angle AEC+\angle AED={180}^{\circ }$, linear pair axiom]
BD =CE [Given]
and AD =AE [Given]
$\therefore \Delta BD\cong \Delta ACE$ [by SAS congruence rule]