Track / Stage 5 / 316 of 400 #916 of 1964

Problem 916

AIME late

Geometry Difficulty 5.8 Prove it

Six. (Full marks 12 points) As shown in the figure, $AB$ is the diameter of a semicircle, $AC \perp AB, AC=AB$ , and any point $D$ is taken on the semicircle. Construct $DE \perp CD$ , intersecting the line $AB$ at point $E$ , and $BF \perp AB$ , intersecting the extension of line segment $AD$ at point $F$ .
(1) If the arc $\overparen{AD}$ is $x^{\circ}$ , to make point $E$ lie on the extension of line segment $BA$ , the range of $x$ is $\qquad$ \_;
(2) Regardless of where point $D$ is taken on the semicircle, apart from $AB = AC$ , there are two other line segments that are always equal. Identify these two equal line segments and provide a proof.

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

$\text { Six, (1) } 0<x<90 \text {. }$
(2) Special positions and the observation of the trend of motion changes can be used to exclude impossible situations from the opposite side, or to conjecture $B E = B F$ from the positive side. The proof is given below.
Connect $B D$ .
$\because A B$ is the diameter of the semicircle, $\therefore B D \perp A D$ .
$\because B F \perp A B, \therefore \triangle B D F \sim \triangle A D B$ .

Thus, $\quad \frac{B F}{A B}=\frac{B D}{A D}$ .
In $\triangle B D E$ and $\triangle A D C$ ,
$\begin{array}{l} \because \angle D B E=\angle D A C, \\ \angle B D E=\angle A D C\left(=90^{\circ}-\angle A D E\right), \\ \therefore \triangle B D E \sim \triangle A D C . \text { Hence } \frac{B E}{A C}=\frac{B D}{A D}, \end{array}$
$\begin{array}{l} \text { Therefore, } \frac{B F}{A \bar{B}}=\frac{B E}{A C} . \\ \because A B=A C, \therefore B E=B F . \end{array}$

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