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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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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

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