# 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.

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

2. $EG = 2\cdot AM$.

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, $AH\perp EG$.