# Midpoints and Orthogonality in Isosceles Triangles

## What is this about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

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 \(AD \perp CH\).

What if applet does not run? |

### 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 \(\triangle MHA\).

## Related material
| |

| |

| |

| |

| |

| |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

62049670 |