Track / Stage 5 / 220 of 400 #820 of 1964

Problem 820

AIME late

Geometry Difficulty 5.6 Find the answer

7. In triangle $ABC$ , the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$ , $ED \parallel AC$ . Find the length of segment $ED$ , given that $AE = 15$ , $AC = 12$ .

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 3.

Fig. 5: to the solution of problem 9.7

Solution. On the ray $AL$ beyond point $L$ , mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$ , it is a parallelogram (in particular, $AC = DX$ ). Therefore, $DX \parallel AC$ . Since $AC \parallel ED$ by the condition, the points $X, D, E$ lie on the same line.

Since $AC \parallel EX$ , then $\angle EAX = \angle CAX = \angle AXE$ , i.e., triangle $AEX$ is isosceles, $EA = EX$ . Then

$ED = EX - XD = EA - AC = 15 - 12 = 3$

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