Track / Stage 6 / 113 of 400 #1113 of 1964

Problem 1113

National olympiad, first round

Geometry Difficulty 6.1 Prove it

8.3. Inside parallelogram $A B C D$ , a point $E$ is taken such that $C E = C B$ . Let $F$ and $G$ be the midpoints of segments $C D$ and $A E$ respectively. Prove that line $F G$ is perpendicular to line $B E$ .

To the solution of problem 8.3

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

Solution: Let $H$ be the midpoint of segment $BE$ . Then $GH$ is the midline of triangle $ABE$ , hence,

$GH = \frac{AB}{2} = \frac{CD}{2} = CF$

and lines $GH$ , $AB$ , and $CD$ are parallel. Therefore, in quadrilateral $HCFG$ , the opposite sides $GH$ and $CF$ are equal and parallel, making this quadrilateral a parallelogram. This means that line $FG$ is parallel to line $CH$ , and it suffices to show the perpendicularity of lines $CH$ and $BE$ . This perpendicularity follows from the fact that triangle $BCE$ is isosceles, so its median $CH$ is also its altitude.

Recommendations for checking:

is in the work	points
Correct proof	7 points
Both geometric constructions (see the point
for 2 points) are considered, but the proof is not completed	5 points
One of the two geometric constructions is considered: 1) midline $GH$ in triangle $ABE$ or 2) median $FH$ in triangle $BFC$	2 points
Any geometric constructions and arguments that do not explicitly lead to the proof	0 points

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