Question

In Fig. BD and CE are two altitudes of a $\Delta$ ABC such that BD = CE.

Prove that the ABC is an isoceles triangle. [2 MARKS]

Answer 1

Process : 1 Mark
Proof : 1 Mark

Proof: In $\Delta$ ABD and ACE, we have
$\angle$ADB = $\angle$AEC = ${90}^{\circ }$
$\angle$BAD = $\angle$CAE [Common angle]
and, BD = CE [Given]
So, by AAS criterion of congruence, we have
$\Delta$ABD $\cong$ $\Delta$ ACE
$\Rightarrow$ AB = AC [ Corresponding parts of congruent triangle are equal]
Hence, $\Delta$ABC is isosceles