Question

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 

Answer 1

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

 

Answer 2

Answer:

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.

Step-by-step explanation: