$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig). Show that
$(i)$ $\Delta ABE \cong \Delta ACF$
$(ii)$ $AB = AC$,i.e.,$ABC$ is an isosceles triangle.
$(i)$ $\Delta ABE \cong \Delta ACF$
$(ii)$ $AB = AC$,i.e.,$ABC$ is an isosceles triangle.
(N/A) $(i)$ In $\Delta ABE$ and $\Delta ACF$,we have:
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.
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