In the adjoining figure, ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || BA. Show that

(i) ∠DAC = ∠BCA

(ii) ABCD is a parallelogram.

(i) In ∆ABC

AB = AC [Given]

∠C = ∠B [Angles opposite to equal sides are equal]

Since, ext. ∠PAC = ∠B + ∠C (Exterior angle is equal to the sum of opposite interior angles)

= ∠C + ∠C

= 2∠C

= 2∠BCA

Since AD bisects ext. ∠PAC, ∠PAC = 2∠DAC

⇒ 2∠DAC = 2∠BCA

⇒ ∠DAC = ∠BCA

Hence, proved that ∠DAC = ∠BCA.

(ii) Since, ∠DAC and ∠BCA are alternate angles and AC is transversal.

It proves that AD || BC.

Since, AD || BC and CD || BA.

Hence, proved that ABCD is a || gm.