Question

In Figure $ \mathrm{D}$ and $ \mathrm{E}$ are points on side $ \mathrm{BC}$ of a $ \mathrm{\Delta ABC}$ such that $ \mathrm{BD}=\mathrm{CE}$ and $ \mathrm{AD}=\mathrm{AE}$. Show that $ \mathrm{\Delta ABD}\cong \mathrm{\Delta ACE}.$

Answer 1

Step 1: Prove $\angle \mathrm{ADB}=\angle \mathrm{AEC}.$

Given: $\mathrm{BD}=\mathrm{CE}$ and $\mathrm{AD}=\mathrm{AE}$

$\mathrm{In}∆\mathrm{ADE},\mathrm{AD}=\mathrm{AE}$ (Given)

Since opposite angles to equal sides are equal.

$\therefore \angle \mathrm{ADE}=\angle \mathrm{AED}-\left(1\right)$

Now,

$\angle \mathrm{ADB}+\angle \mathrm{ADE}=180°$ (Linear Pair)

$\angle \mathrm{ADB}=180°-\angle \mathrm{ADE}-\left(2\right)$

Similarly,

$\angle \mathrm{AEC}=180°-\angle \mathrm{AED}-\left(3\right)$

From (1) and (2)

$\angle \mathrm{ADB}=180°-\angle \mathrm{AED}-\left(4\right)$

From (3) and (4)

$\angle \mathrm{ADB}=\angle \mathrm{AEC}$

Step 2: Prove $\mathrm{\Delta ABD}\cong \mathrm{\Delta ACE}.$

$\mathrm{In}\mathrm{\Delta ABD}\mathrm{and}\mathrm{\Delta ACE}$,

$\mathrm{AD}=\mathrm{AE}$ (Given)

$\angle \mathrm{ADB}=\angle \mathrm{AEC}$ (From Step 1)

$\mathrm{BD}=\mathrm{EC}$ (Given)

Hence $\mathrm{\Delta ABD}\cong \mathrm{\Delta ACE}$ (by SAS congruency)

Hence proved $\mathrm{\Delta ABD}\cong \mathrm{\Delta ACE}.$