Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

Let ABC be an isosceles triangle with AB = AC.

CD and BE are perpendiculars drawn from extremities of base BC.

In △ADC and △AEB,

⇒ AC = AB (Given)

⇒ ∠AEB = ∠ADC (Both equal to 90°)

⇒ ∠A = ∠A (Common angle)

∴ △ADC ≅ △AEB (By A.A.S axiom)

⇒ BE = CD (Corresponding parts of congruent triangles are equal)

Hence, the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.