# Square, Similarity and Slopes

The following problem was twitted around by David Radcliffe:

$ABCD$ is a square. Point $E$ is on side $AB$ or its extension. $F$ is the intersection of $DE$ and $BC$. $G$ is the intersection of $AF$ and $CE$. Prove that $BG$ is perpendicular to $DE$.

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Solution $ABCD$ is a square. Point $E$ is on side $AB$ or its extension. $F$ is the intersection of $DE$ and $BC$. $G$ is the intersection of $AF$ and $CE$. Prove that $BG$ is perpendicular to $DE$.

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### Solution

Depending of where point $E$ is relative to $AB$ there are three possible configurations. The same idea applies in all three cases. I shall consider only one where point $E$ is to the right from $B$. Extend $AG$ and $BG$ to their intersection with $CD$ at $L$ and $M$, respectively. This generates several pairs of similar triangles. I'll use the similarities to determine the slopes of $BG$ and $DE$. To start with,

\begin{align} slope(DE) & = -AD/AE, \\ slope(BG) & = BC/CM. \end{align}

Triangles $CGM$ and $EGB$ are similar and so are triangles $LGM$ and $AGB$. From here,

$\frac{CM}{BE} = \frac{GM}{BG} = \frac{LM}{AB}.$

Triangles $DCF$ and $EBF$ are also similar and so are triangles $CLF$ and $BAF$, so that

$\frac{CL}{AB} = \frac{CF}{BF} = \frac{CD}{BE} = \frac{AB}{BE}.$

Or,

$\frac{AB}{BE} = \frac{CM + LM}{AB} = \frac{CM}{AB} + \frac{CM}{BE}.$

This can be rewritten as

$AB^2 = CM(AB + BE) = CM\cdot AE.$

And, finally,

$1 = \frac{AB}{CM} \cdot \frac{AB}{AE} = \frac{BC}{CM} \cdot \frac{AD}{AE},$

which is the same as

$slope(DE)\cdot slope(BG) = -1,$

implying that the two lines are orthogonal. • 