Question

$ABC$ is triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal. How many of the following are true:
(a) $\Delta ABE\cong \Delta ACF$
(b) $AB=AC$, i.e., is an isosceles triangle.

Answer 1

$GivenBE=CF-\left(1\right)$

$BE\&CFarealtitudes$

$so\angle AEB={90}^{\circ }and\angle AFC={90}^{\circ }-\left(2\right)$

$toshow:\Delta ABE\cong \Delta ACF\&$

$AB=AC$

$In\Delta ABEand\Delta ACF$

$\angle AEB=\angle AFC$

$\angle A=\angle A\left[from\left(2\right)\right]$

$BE=CFfrom\left(1\right)$

$\Delta ABE\cong \Delta ACF\left(AAScongruencerule\right)$

$AB=AC\left(CDCT\right)$