Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 4.5 AIME Prove it Brazil

$ABCD$ is any convex quadrilateral. Squares center $E, F, G, H$ are constructed on the outside of the edges $AB, BC, CD$ and $DA$ respectively. Show that $EG$ and $FH$ are equal and perpendicular.

Solution

Consider $A, B, C, D$ to be the points $a, b, c, d$ in the complex plane. Then $E, F, G, H$ are the points $\frac{a+b}{2} + i\frac{b-a}{2}$ , $\frac{b+c}{2} + i\frac{c-b}{2}$ , $\frac{c+d}{2} + i\frac{d-c}{2}$ , $\frac{d+a}{2} + i\frac{a-d}{2}$ . Hence vector $\vec{GE}$ is represented by the complex number $h = \frac{a+b-c-d}{2} + i\frac{-a+b+c-d}{2}$ and the vector $\vec{FH}$ by the $k = \frac{a-b-c+d}{2} + i\frac{a+b-c-d}{2}$ . But $k = ih$ , so $EG$ and $FH$ are equal and perpendicular.

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