Show that a quadrilateral is a square (2015 OMCS #4)

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High School Math Grades 9-12, Ages 13-18

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Show that a quadrilateral is a square (2015 OMCS #4)

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#1 h May 16, 2015, 11:22 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a convex quadrilateral such that $\angle{BAD} = 90^{\circ}$ and its diagonals $AC$ and $BD$ are perpendicular. Let $M$ be the midpoint of side $CD$ , and $E$ be the intersection of $BM$ and $AC$ . Let $F$ be a point on side $AD$ such that $BM$ and $EF$ are perpendicular. If $CE = AF\sqrt{2}$ and $FD = CE\sqrt{2}$ , show that $ABCD$ is a square.

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#2 h May 17, 2015, 12:32 AM • 3 Y

Y by FabrizioFelen, Adventure10, Mango247

I have the test, too, this level is embarrassment; but is more hard than the problem 3. Latin american(olympic team) is bad in geometry $\ldots$
WLOG $EC=\sqrt{2},FD=2,AF=1$
$BD \cap AC \equiv X$ By Menelaus' theorem ( $\triangle XCD - M,E,B$ ) we get $\frac{\sqrt{2}}{EX}=\frac{BD}{BX}$
Since $B$ is the spiral-center $AF \rightarrow XE \Rightarrow \frac{AB}{1}=\frac{BX}{XE} \Rightarrow BD=AB\sqrt{2} \Rightarrow \angle ADB=45^{\circ}$ and $3XE=BX=\frac{3\sqrt{2}}{2}=EC+XE \Rightarrow AX=BX=CX=DX \ldots$

This post has been edited 3 times. Last edited by drmzjoseph, May 17, 2015, 1:04 AM
Reason: My English is bad :(

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