Given: Triangle ABC is isosceles with AB = AC. Side AC is extended to point D in such a way that C is between A and D (but not necessarily the midpoint). Then E is found between B and C so that EC = CD. DE is now extended to intersect AB at F.
To Prove: angle AFE is thrice the angle at D.
Solution
Let angle D = . Then angle E1 = x and angle C1 = (exterior angle theorem). Then angle B = and angle F1 = (exterior angle again).
QED.
Which stands for "Quod Erat Demonstrandum", meaning "that which was to be proved." (Students who try to get away with "Quite Easily Done" should be stared at, hard, while muttering.)
