Question

In parallelogram ABCD, AC is a diagonal. △ABC is an isosceles triangle where ACB > ACD, ACD = x° and BCD = 110°. What is the value of x?

Answer 1

Given : ABC is an isosceles triagle in which AB=AC.AD bisects exterior angle QAC and CD∣∣BA.

To show : 

(i)∠DAC=∠BCA

(ii) ABCDABCD is a parallelogram

Proof :

(i)

∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)

∠QAD=∠DAC=x(let) (Given)

∠DCA=∠BAC=z(let) (Alternate interior angles)

And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.

So,

∠QAD+∠DAC=∠ABC+∠BCA

x+x=y+y

2x=2y

x=y

∠DAC=∠BCA (hence proved)

(ii)

Now because,

∠DAC=∠BCA (proved above)

Therefore , AD∣∣BC

And CD∣∣BA (Given)

Since opposite sides of quadrilateral ABCD are parallel therefore ABCD is a parallelogram.

Hope This is helpful for you