# Two Properties of Flank Triangles - and a Third One

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.

### Proof

For a proof, complete parallelograms $AGIE$ and $ABJC$. The two parallelograms are equal, since angles $BAC$ and $EAG$ are supplementary, and the sides in both are those of the two squares. $AM$ is just a half of the diagonal; $EG$ is the corresponding whole diagonal in another parallelogram.

To prove that $AJ\perp EG$ suffice it to observe a pair of equal triangles with the corresponding sides at right angles. A pair $\triangle ABJ$ and $\triangle AEG$ serves that purpose.

The configuration may be expanded with four line segments that form a square, $DJFI$.

That this is indeed a square follows from the fact that the four triangles $BDJ$, $CJF$, $GIF$, and $EDI$ are equal. Also, the angles in the rhombus $DJFI$ are all right. For example, $\angle IDJ = \angle BDE -\angle EDI + \angle BDJ = 90^{\circ}$, because $\angle BDE = 90^{\circ}$ while $\angle EDI = \angle BD$.

This supplies an additional proof to an observation of the late Professor McWorter to the effect that $\angle DJF$ is right.

In passing, the center of square $DJFI$ is of course the point which is the subject of Bottema's Theorem.

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