Question

In triangle ABC AB equal to AC and angle C A D is an exterior angle. If ray AP is the bisector of angle C A D then prove that AP parallel to BC

Answer 1

Answer:

Since, AE∥BC,

thus, ∠EAC=∠ACB (Alternate angles)

Also, ∠DAE=∠EAC (AE bisects ∠ DAC)

∠DAC=∠ACB+∠ABC (Exterior angle is equal to sum of interior opposite angles)

∠ABC=∠DAC−∠ACB

or ∠ABC=∠DAC−∠EAC

∠ABC=∠DAE=∠EAC=∠ACB

Now, In △ABC

Since, ∠ABC=∠ACB

Hence, AB=AC (Opposite sides of equal opposite angles are equal)