# Two Properties of Flank Triangles - First Proof by Symmetric Rotation

Bottema's configuration of two squares that share a vertex is naturally embedded into Vecten's configuration of three squares erected on the sides of a triangle. The latter generalizes the first of Euclid's proofs of the Pythagorean theorem, so it rightfully refers to the *Bride's Chair*.

In the discussion of Vecten's configuration, we proved *inter alia* two properties of flank triangles that are a part of Bottema's configuration.

If \(AM\) is a median in \(\triangle ABC\), \(AH\) is an altitude in \(\triangle AEG\), then \(M\), \(A\), \(H\) are collinear.

\(EG = 2\cdot AM\).

The applet below illustrates one of the solutions. Use the slider at the top of the applet to rotate \(\triangle AEG\) \(90^{\circ}\) clockwise around \(A\).

### Proof

The rotation of \(\triangle AEG\) \(90^{\circ}\) clockwise maps it on \(\triangle AE'C\); and since \(\angle BAE=90^{\circ}\), points \(B\), \(A\), and \(E'\) are collinear. In \(\triangle BCE'\), \(M\) is the midpoint of side \(BC\), \(A\) is the midpoint of side \(BE'\), so that \(AM\) as a midline is parallel to \(CE'\) and equals half its length. But \(CE'=EG\) which proves the second property.

For a proof of the first property, observe that, since \(AM\parallel CE'\) and \(CE'\perp EG\) (by the construction), \(AM\perp EG\), but \(AH\) is on the same line as \(AM\). Therefore, \(AH\perp EG\).

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