## A theorem of two intersecting circles or two adjacent isosceles triangles

(An application of a generalization of Van Aubel's Theorem)

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The following intriguing problem was circulated by Dick Klingens at the annual meeting of the Nederlandse Vereniging van Wiskundeleraren (NVvW) in November 2011:

Two circles *K*_{1} and *K*_{2} intersect each other in *B* and *Q*. A line through *Q* cuts *K*_{1} in point *A* and *K*_{2} in point *C*. The points *E* and *F* are the respective midpoints of the arcs *BA* and *BC* that do not contain *Q*. If *M* is the midpoint of *AC*, prove that ∠*EMF* = 90^{o}

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Dick Klingens' Problem

A different, but equivalent formulation of the same problem is the following:

Given two isosceles triangles *ABC* and *BDE* with respective bases *AB* and *BD* lying in a straight line (and on the same side of the line), and *BC* is perpendicular to *BE*. If *M* is the midpoint of *AD*, prove that ∠*CME* = 90^{o}.

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Alternative, equivalent formulation

**Challenge**: Can you prove either one of these formulations? This extremely rich problem can be proved in many different ways, which enhances its beauty and intrigue. If you get stuck, one possible hint is given in the second version above - press the *Hint* button in the sketch.

Compare your **own** solution to this one, *'Application of a generalization of Van Aubel'*, which includes a clickable link to another interactive sketch as well as a clickable link to a paper in Dutch with 5 different proofs of the result, plus a download link to 9 more different proofs. Just recently in Sept 2013, my friend and colleague Poobhal Pillay, the Siyanqoba Coordinator for KZN, supplied the following short, elegant proof of the 2nd version above using only the theorem of Pythagoras: *'Poobhal Proof'*

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By Michael de Villiers. Created, 18 August 2013.