Question

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Answer 1

Given:

• An isosceles triangle ABC in which AB = AC .

• AD is the bisector of exterior ∠PAC.

• CD | | AB

To Prove:

∠DAC = ∠BCA and ABCD is a parallelogram.

Proof:

In ∆ABC

AB = AC [ Given ]

∠B = ∠C [ Angles opp. to equal sides ]

In ∆ABC

Exterior angles = Sum of two opposite interior angles

∠CAP = ∠1 + ∠2

⇒ ∠CAP = 2∠2

★AD is the bisector of ext. ∠ CAP .

Therefore, ∠CAP = 2∠3

⇒ 2∠3 = 2∠2

⇒ ∠3 = ∠2

⇒ ∠DAC = ∠BCA

Now,

AC intersect lines AD and BC at A and C respectively.

∠3 = ∠2 [ alternate interior angles ]

Therefore, AD | | BC

But , CD | | AB [ Given ]

Thus,the ABCD is parallelogram [ AD | | BC and CD | | AB ]