Question

In the following figure, D and E are points on the base BC of a triangle ABC such that BD = CE and AD = AE. Prove that Δ ABE $\cong$ Δ ACD.

Answer 1

Since AD = AE, so $∆ADE$ is a an isosceles triangle and therefore
$$\begin{gathered} \angle ADE=\angle AED \\ \Rightarrow {180}^{\circ }-\angle ADE={180}^{\circ }-\angle AED \\ \Rightarrow \angle BDA=\angle AEC.....\left(i\right) \end{gathered}$$
Now, in $∆ABD\mathrm{and}∆ACE$, we have
$$\begin{gathered} BD=EC\left(\mathrm{Given}\right) \\ AD=AE\left(\mathrm{Given}\right) \\ \angle BDA=\angle AEC\left[\mathrm{From}\left(i\right)\right] \\ \therefore ∆ABD\cong ∆ACE\left(\mathrm{By}\mathrm{SAS}\right) \end{gathered}$$
$\Rightarrow AB=AC\left(\mathrm{BY}\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}\right)$
In $∆ABE\mathrm{and}∆ACD$
$$\begin{gathered} BD=EC\left(\mathrm{Given}\right) \\ \Rightarrow BD+ED=EC+ED \\ \Rightarrow BE=CD \end{gathered}$$
AB = AC
AE = AD
$\therefore ∆ABE=∆ACD\left(\mathrm{By}\mathrm{SSS}\right)$
Hence, Δ ABE $\cong$ Δ ACD.