Question

ABC is an isosceles triangle in which ab is equal to AC and bisects exterior angle DAC and CD parallel AB show that DAC is equal to BCA and ABCD is a parallelogram

Answer 1

Given : AB = AC

CD ║AB

To prove : ΔBCA ≅ ΔDAC

and ABCD is parallelogram , i.e, One pair of opposite arms is parallel and equal. So, it is enough to prove AB = CD.

Proof:

Consider ΔDAC and ΔBCA

∠DAC = ∠BCA (Alternate angles)

AC = CA (Common)

∠DCA = ∠BAC (Alternate angles)

∴ΔDAC ≅ ΔBCA by ASA Congruence condition.

Since the triangles are congruent, we have

BC = AD ,

AB = CD and

∠ABC = ∠ADC

Since, AB = CD and AB ║CD, ABCD is a parallelogram.

Hence proved.

Answer 2

Answer:

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

Step-by-step explanation: