Midpoints and Orthogonality in Isosceles Triangles

What is this about? A Mathematical Droodle

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Solution

The applet was supposed to remind of a problem from an oldish Russian collection:

In isosceles triangle $ABC~(AB = AC)$, $M$ is the midpoint of $BC$, $MH \perp AB$, and $D$ is the midpoint of $HM$. Prove that $AD \perp CH$.

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Solution

For a proof, let $CL$ be perpendicular to $AB$. Triangles $BLC$ and $BMA$ are both right and share angle at $B$, they are therefore similar: $\triangle BLC\sim\triangle BMA$. For the same reason, $\triangle BMA\sim\triangle MHA$, so that $\triangle BLC\sim\triangle MHA$. In the two triangles all pairs of corresponding sides are perpendicular: $CL\perp AH, BL\perp MH, CH\perp AM$. It follows that the two triangles are obtained from each other by a spiral similarity that includes a rotation by $90^{\circ}$. It follows that all the corresponding elements of the two triangles are perpendicular. In particular, this is true of $CH$ and $AD$ which are the medians to the sides $BL$ of $\triangle BLC$ and that of $AD$ in $\triangle MHA$.

• Orthogonality in Isosceles Triangles
• A Median in Touching Circles
• Two Altitudes, One Midpoint
• Square, Similarity and Slopes