Question

In a ΔABC, the internal bisector of angle A meets the opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that ΔACE is isosceles.

Answer 1

DA || CE                 ... [Given]
⇒ ∠1 = ∠4            ...(i) ( Corresponding angles )
∠2 = ∠3                ....(ii) ( Alternate angles )
But ∠1 = ∠2          ....(iii) ( AD is the bisector of ∠A )

From (i), (ii) and (iii)
∠3 = ∠4
⇒ AC = AE
⇒ ΔACE is an isosceles triangle.