# Properties of Flank Triangles - Second 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 \(\triangle AEG\), \(AH\) is an altitude in \(\triangle ABC\), then \(M\), \(A\), \(H\) are collinear.

\(BC = 2\cdot AM\).

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

So we rotate \(\triangle AEG\) \(90^{\circ}\) degrees clockwise; after the rotation, \(AM'\perp AM\); \(E\) maps onto \(E'\), \(M\) onto \(M'\), \(G\) onto \(C\); \(A\), \(B\), \(E'\) are collinear; \(A\) the midpoint of \(BE'\); \(M'\) the midpoint of \(CE'\). It follows that \(AM'\) is the midline in \(\triangle BCE'\).

Therefore, \(BC = 2\cdot AM'\ = 2\cdot AM\). Also \(AM'\parallel BC\), implying \(AM\perp BC\). In other words, \(M\), \(A\), \(H\) are collinear.

[Honsberger, p 65] actually proves that if \(M\) lies on the continuation of the altitude \(AH\) then it's the midpoint of \(EG\). I leave this as an exercise.

### References

- R. Honsberger,
*Mathematical Diamonds*, MAA, 2003

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