Question

Let AB'C be a triangle with ∠B'AC=70∘, ∠ACB'=80∘, F is a point inside the triangle such that ∠FB'A=∠FCB‘=20∘. Find ∠AFB'.

Answer 1

Answer:

In ΔABE and ΔACF,

∠BAE=∠CAF (Common angle)

∠AEB=∠AFC ....(∵BE⊥AC and CF⊥AB)

BE=CF (Given that altitudes are equal)

By AAS criterion of congruence,

ΔABE≅ΔACF

Hence,

AB=AC