Question

In the adjoining fig. ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD || AB. Show that ABCD is a parallelogram

Answer 1

plz look at the image...if you still don't understand please message me...I ll tell easy method to do it

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: