(A variation of Reim's theorem)

"*The definition of a good mathematical problem is the mathematics it generates rather than the problem itself*." — Andrew Wiles from an interview for PBS website on the NOVA program, 'The Proof'.

The following exploration is generalized from a question in the Mathematics Paper 2 of the 1949 National Senior Examinations for Grade 12 for the (then) Union of South Africa.

**Matric Problem (Given)**

Given any trapezium *ABCD* with *AB* // *DC* and line *EF* constructed as indicated so that ∠*AEF* = ∠*DCF*.

**Exploration**

1) Click on the 'Show Angles' button to display the angle measurements of ∠'s *BEC* and *AFD*. What do you notice?

2) Explore your observation in 1) above by dragging any of *A*, *B*, *C*, *D* or *E*. Can you deductively explain (prove) why your observation is true?

3) Is the result still valid if *E* is dragged 'outside' *AD* onto its extensions on both sides? What if point *D* is dragged towards *C*, and then past *C*? Is the result still valid?^{1}

^{1} Note that when point *D* is dragged past *C*, the given condition becomes ∠*AEF* = 180° - ∠*DCF*.

4) **Challenge**: Can you deductively explain (prove) your observations above?

Matric Exam Geometry Problem - 1949

**Further exploration**

5) Can you formulate alternative, equivalent versions of the result? Can you create your own dynamic geometry sketch?

6) When, or rather where, is ∠*BEC* a maximum as *E* is dragged along line *AD*? Can you determine the optimal position?

*Hint*: If you get stuck with 6) above, go to this dynamic sketch Determining maximum angle. Or use it to check & compare your solution.

7) Recently (Jan 2024) on my Facebook page, colleague Thanos Kalogerakis from Kiáto, Greece suggested another variation, which is shown below.

*Hint*: If you get stuck with 7) above, go to this link Thanos solution - construction of EF.

**Published Paper**

Check your explorations for 1)-6) above, by reading my 2015 paper in *At Right Angles* and *Learning and Teaching Mathematics* at *"Flashback to the Past: a 1949 Matric Geometry Question"*.

**Reim's theorem**

Thanos Kalogerakis also kindly brought to my attention that the basic result mentioned at the start is a variation of what is apparently known as Reim's theorem. It is named after Anton Reim (1832–1922), a Czech mathematician. Read more about this theorem at the following links:

Art of Problem Solving: Reim's Theorem

Art of Problem Solving: Lemmas in AoPS Geometry by Evan Chen

Art of Problem Solving: Used in solution for IMOC 2019 G5

Cut the Knot: Reim's Similar Coins

Maths Olympian: Reim's theorem

Pleasanton Math Circle: Notes on Cyclic Quadrilaterals.

**Note**: If anyone knows more about the origin of this interesting theorem or applications of it, please let me know.

**Related Links**

Matric Exam Geometry Problem - 1949: Maximizing ∠BEC

All parabola are similar - i.e. have the same shape

Cyclic Hexagon Alternate Angles Sum Theorem

A generalization of the Cyclic Quadrilateral Angle Sum theorem

Angle Divider Theorem for a Cyclic Quadrilateral

Semi-regular Angle-gons and Side-gons: Generalizations of rectangles and rhombi

Alternate sides cyclic-2n-gons and Alternate angles circum-2n-gons

Nine-point centre (anticentre or Euler centre) & Maltitudes of Cyclic Quadrilateral

An extension of the IMO 2014 Problem 4

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Created by Michael de Villiers, 13 August 2015; updated to *WebSketchpad* 18 Jan 2024, updated 25 Jan 2024.