Track / Stage 4 / 77 of 340 #337 of 2000

Problem 337

AMC 12 late, AIME early

Geometry Difficulty 4.6 Find the answer

1. In trapezoid $A B C D$ , it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$ . If $A E=10$ , then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition)

Your answer

Official solution

(提示: Extend $D A$ to point $F$ , complete the trapezoid into a square $C B F D$ , then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$ . It is easy to see that $\triangle A B E \cong \triangle G B E$ . Therefore, $A E$ $=E G=E C+A F=10$ . Let $E C=x$ , then $A F=10-$ $x, D E=12-x, A D=2+x$ . By the Pythagorean theorem, $(12-x)^{2}+(2+x)^{2}=10^{2}$ . Solving this, we get $x=4$ or 6. )

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