Olympiad Maths Prep

Track / Stage 4 / 69 of 340 #329 of 2000

Problem 329

AMC 12 late, AIME early
Geometry Difficulty 4.6 Find the answer

Example 4 As shown in Figure 4, in rhombus ABCDA B C D, ABC\angle A B C =120,F=120^{\circ}, F is the midpoint of DCD C, and the extension of AFA F intersects the extension of BCB C at point EE. Then the degree measure of the acute angle formed by line BFB F and DED E is ( ).
(A) 3030^{\circ}
(B) 4040^{\circ}
(C) 5050^{\circ}
(D) 6060^{\circ}
(2008, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition)

Official solution

As shown in Figure 4, let BFB F intersect DED E at point MM, and connect BDB D. Then BCD\triangle B C D is an equilateral triangle.
From FF being the midpoint of CDC D, we know MBC=30\angle M B C=30^{\circ}.
Since AD//CEA D / / C E, then ADF\triangle A D F and ECF\triangle E C F are centrally symmetric about point
FF. Therefore, CE=AD=CDC E=A D=C D.
Thus, CEM=30,DMF=60\angle C E M=30^{\circ}, \angle D M F=60^{\circ}.

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