$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using $RHS$ congruence rule,prove that the triangle $ABC$ is isosceles.

(N/A) Given: $BE \perp AC$ and $CF \perp AB$ such that $BE = CF$.
To prove: $\Delta ABC$ is an isosceles triangle,i.e.,$AB = AC$.
Proof:
In right-angled triangles $\Delta BEC$ and $\Delta CFB$:
$1$. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes are perpendicular to the sides).
$2$. $BC = CB$ (Common hypotenuse).
$3$. $BE = CF$ (Given).
Therefore,by $RHS$ congruence rule,$\Delta BEC \cong \Delta CFB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$\angle BCE = \angle CBF$
This implies $\angle BCA = \angle CBA$.
In $\Delta ABC$,since the angles opposite to equal sides are equal,the sides opposite to equal angles are also equal.
Therefore,$AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.