# In the given figure, BE and CF are two equal altitudes of ΔABC.

Question:

In the given figure, BE and CF are two equal altitudes of ΔABC.
Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Solution:

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACF}$, we have :

$\mathrm{BE}=\mathrm{CF} \quad$ (Given)

$\angle \mathrm{BEA}=\angle \mathrm{CFA}=90^{\circ}$

$\angle \mathrm{A}=\angle \mathrm{A}$ (Common)

$\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF} \quad$ (AAS criterion)

$\mathrm{AB}=\mathrm{AC}(\mathrm{CPCT})$