Question

$\text{ABC}$ is a triangle in which altitudes $\text{BD}$ and $\text{CE}$ to sides $\text{AC}$ and $\text{AB}$ are equal (see figure). Show that $\text{AB = AC}$ i.e. $\text{ABC}$ is an isosceles triangle

Answer 1

Given ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal

In $\Delta ADB$ and $\Delta ACE$

$\angle ADB=\angle AEC={90}^0$ (altitudes BD and CE to sides AC and AB)

$BD=CE$ (Given altitudes are equal)

$\angle BAD=\angle CAE$ (Common angle )

$\Delta ABD\cong \Delta ACE$ [by $AAS$ congruence]

So $AB=AC$ (by CPCT)

$\therefore$ triangle $ABC$ is an isosceles triangle