Equal sides AB and AC of an isosceles triangle ABC are produced. The bisectors of the exterior angles so formed meet at D. Prove that AD bisects angle A.

ABC is an isosceles triangle and AB is produced to E and AC is produced to F. BD and CD are bisectors of angle CBE and angle BCF respectively.

In Δ ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides are equal)

Since, ABE is a straight line.

∴ ∠ABC + ∠CBE = 180°

⇒ ∠CBE = 180° - ∠ABC

Since, BD is bisector of ∠CBE,

$\therefore ∠CBD = \dfrac{∠CBE}{2} \\[1em] = \dfrac{180° - ∠ABC}{2} \\[1em] = 90° - \dfrac{∠ABC}{2} \\[1em] = 90° - \dfrac{∠ACB}{2} \space ….(1)$

Since, ACF is a straight line.

∴ ∠BCF + ∠ACB = 180°

⇒ ∠BCF = 180° - ∠ACB

Since, CD is bisector of ∠BCF,

$\therefore ∠BCD = \dfrac{∠BCF}{2} \\[1em] = \dfrac{180° - ∠ACB}{2} \\[1em] = 90° - \dfrac{∠ACB}{2} \space ….(2)$

From equations (1) and (2), we get :

⇒ ∠CBD = ∠BCD.

In △ BCD,

⇒ BD = CD (Sides opposite to equal angles are equal)

In △ ABD and △ ACD,

⇒ AB = AC (Given)

⇒ BD = CD (Proved)

⇒ AD = AD (Common side)

∴ △ ABD ≅ △ ACD (By S.S.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠BAD = ∠CAD (By C.P.C.T.C.)

Hence, proved that AD bisects ∠A.