Track / Stage 5 / 304 of 400 #904 of 1964

Problem 904

AIME late

Geometry Difficulty 5.8 Prove it

Points $D, E$ , and $F$ are chosen on the sides $A C, A B$ , and $B C$ of isosceles triangle $A B C (A B=B C)$ such that $D E = D F$ and $\angle B A C = \angle F D E$ .

Prove that $A E + F C = A C$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $\angle A=\angle C=\alpha, \angle A D E=\beta$ . Triangles $A E D$ and $C D F$ are congruent by side and two adjacent angles, since $D E=D F, \angle D A E=\angle D C F, \angle A E D=180^{\circ}-\alpha-\beta=\angle C D F$ , therefore $A E=C D$ and $A D=C F$ . Consequently, $A E+F C=C D+A D=A C$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.