$\Delta ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively. Show that these altitudes are equal.
(N/A) Given: $\Delta ABC$ is an isosceles triangle with $AB = AC$.
Altitudes $BE \perp AC$ and $CF \perp AB$.
To prove: $BE = CF$.
Proof:
In $\Delta ABE$ and $\Delta ACF$:
$1$. $\angle A = \angle A$ (Common angle)
$2$. $AB = AC$ (Given)
$3$. $\angle AEB = \angle AFC = 90^\circ$ (Since $BE$ and $CF$ are altitudes)
Therefore,by $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$BE = CF$.
Altitudes $BE \perp AC$ and $CF \perp AB$.
To prove: $BE = CF$.
Proof:
In $\Delta ABE$ and $\Delta ACF$:
$1$. $\angle A = \angle A$ (Common angle)
$2$. $AB = AC$ (Given)
$3$. $\angle AEB = \angle AFC = 90^\circ$ (Since $BE$ and $CF$ are altitudes)
Therefore,by $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$BE = CF$.
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