If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.

Let ABC be a triangle.

Let AD be a perpendicular from vertex A to base BC, which bisects it, i.e. DB = DC.

In △ADB and △ADC,

⇒ AD = AD (Common side)

⇒ BD = DC (Given)

⇒ ∠ADB = ∠ADC (Each equal to 90°)

∴ △ADB ≅ △ADC (By S.A.S. axiom)

∴ AB = AC (Corresponding parts of congruent triangles are equal)

∴ Triangle ABC is an isosceles triangle.

Hence, proved that the triangle is isosceles, if the altitude from one vertex of a triangle bisects the opposite side.