## Klingens' theorem of two intersecting circles or two adjacent isosceles triangles(An application of a generalization of Van Aubel's Theorem)

In memoriam: A tribute to Dick Klingens (7 Mei 1945 - 24 Mei 2021)

The following intriguing problem was circulated by Dick Klingens at the annual meeting of the Nederlandse Vereniging van Wiskundeleraren (NVvW) in November 2011:
Formulation 1: Two circles K1 and K2 intersect each other in B and Q. A line through Q cuts K1 in point A and K2 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 = 90o

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Dick Klingens' Problem

A different, but equivalent formulation of the same problem is the following:
Formuluation 2: 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 = 90o.

Important: To view & manipulate the dynamic version of this 2nd formulation, navigate to it using the appropriate button in the ABOVE dynamic sketch; the picture below is static.

#### Alternative, equivalent formulation

Challenge: Can you prove either one of these formulations? This extremely rich problem (but deceptively hard 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 dynamic sketch at the top.

Compare your own solution to this one, 'Application of a generalization of Van Aubel', which includes a link to an interactive sketch as well as a link to a paper in Dutch with 5 different proofs of the result.