# If ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC & CD parallel to AB. show that i) DAZ = LBAC ii) ABCD is a parallelogram

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by monu1234

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by monu1234

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Let BAC be Ф. in ABC, B = C as isosceles Δ.

ABC = BCA = (180 - Ф)/2 = 90-Ф/2

As AB parallel to CD, BAC = ACD = Ф.

Also ACB = CAD = 90 - Ф/2

In ΔCAD, ADC = 180 - ACD - CAD = 180 - Ф - 90 + Ф/2 = 90 -Ф/2

So ADC = ABC.

Opposite angles are same and two sides are parallel, it is a parallelogram

ABC = BCA = (180 - Ф)/2 = 90-Ф/2

As AB parallel to CD, BAC = ACD = Ф.

Also ACB = CAD = 90 - Ф/2

In ΔCAD, ADC = 180 - ACD - CAD = 180 - Ф - 90 + Ф/2 = 90 -Ф/2

So ADC = ABC.

Opposite angles are same and two sides are parallel, it is a parallelogram