Track / Stage 5 / 339 of 400 #939 of 1964

Problem 939

AIME late

Geometry Difficulty 5.7 Prove it

As shown in Figure 6, in square $A B C D$ , $E$ and $F$ are points on sides $B C$ and $C D$ respectively, and $E F=B E+D F$ . $A E$ and $A F$ intersect the diagonal $B D$ at points $M$ and $N$ respectively. Prove:
(1) $\angle E A F=45^{\circ}$ ;
(2) $M N^{2}=B M^{2}+D N^{2}$
【Analysis】This is a

very typical problem of a "right-angled triangle containing $45^{\circ}$ in a square". The problem can be solved through rotation transformation.

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

Proof (1) As shown in Figure 6, rotate $\triangle ABC$ $90^{\circ}$ around point $A$ to get $\triangle ADE_{1}$ .
Thus, $AE=AE_{1}, \angle EAE_{1}=90^{\circ}$ .
Also, $EF=BE+DF=E_{1}D+DF=E_{1}F$ , so $\triangle AEF \cong \triangle AE_{1}F$ .
Therefore, $\angle EAF=\angle E_{1}AF=\frac{1}{2} \angle EAE_{1}=45^{\circ}$ .
(2) Take a point $M_{1}$ on $AE_{1}$ such that $AM_{1}=AM$ , and connect $M_{1}D, M_{1}N$ . Then
$\triangle ABM \cong \triangle ADM_{1}, \triangle ANM \cong \triangle ANM_{1}$ .
Thus, $\angle ABM=\angle ADM_{1}$ ,
$BM=DM_{1}, MN=M_{1}N$ .
Also, $\angle NDM_{1}=90^{\circ}$ , so
$M_{1}N^{2}=M_{1}D^{2}+ND^{2}$ .
Therefore, $MN^{2}=BM^{2}+DN^{2}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.