Prove that the bisectors of the base angles of an isosceles triangle are equal.

Let ABC be an isosceles triangle with AB = AC.

⇒ ∠B = ∠C (Angles opposite to equal sides in a triangle are equal)

CE and BD are the bisectors of angles ∠C and ∠B respectively to sides AB and AC respectively.

⇒ ∠ABD = ∠DBC and ∠ACE = ∠ECB

Since, angles B and C are equal, thus their half will also be equal.

⇒ ∠ABD = ∠DBC = ∠ACE = ∠ECB

In △AEC and △ADB,

⇒ AC = AB (Given)

⇒ ∠A = ∠A (Common angle)

⇒ ∠ACE = ∠ABD (Proved above)

∴ △AEC ≅ △ADB (By A.S.A axiom)

⇒ CE = BD (Corresponding parts of congruent triangles are equal.)

Hence, the bisectors of the base angles of an isosceles triangle are equal.