ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:

(i)    ∆ABE ≅ ∆ACF
(ii)    AB = AC, i.e., ∆ABC is an isosceles triangle.

∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∆BCD is a right angle. 

ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
[Hint. Take a point P on AC such that BP bisects ∠B. Join P and D.]