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

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Copyright © 1996-2017 Alexander Bogomolny

\(ABCD\) 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 \(B\).

Problem in Square: Similarity and Slopes

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.

Related material
Read more...

  • Orthogonality in Isosceles Triangles
  • Midpoints and Orthogonality in Isosceles Triangles
  • A Median in Touching Circles
  • Two Altitudes, One Midpoint
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