## Outline Mathematics

Geometry

# Triangle Areas in a Parallelogram II

This is a generalization of a one hundred years old problem.

If from any point in a diagonal of a parallelogram lines are be drawn to the opposite angles the parallelogram will be divided into two pairs of equivalent triangles.

The problem appeared in a periodical publication *Mathematical Visitor* and then included into a recent collection under the same name. It took the broadcast power of internet for the problem to reach overseas and a generalization emerge. Floor van Lamoen, The Netherlands, has noticed that point M does not have to lie on a diagonal of the parallelogram for there to be two pairs of triangles with equal sum of areas.

So let M be any point inside parallelogram ABCD. Triangles AMB and CMD have equal bases:

Area(AMB) + Area(CMD) | = AB·ME/2 + CD·MF/2 |

= AB·ME/2 + AB·MF/2 | |

= AB·(ME + MF)/2 | |

= AB·EF/2 | |

= Area(ABCD)/2,Area(ABC)/2,Area(ABCD),Area(ABCD)/2. |

It follows that the remaining area,

Matt Henderson came up with an even simpler solution:

The two lines through point M parallel to the sides of the parallelogram cut the latter into four smaller parallelograms. Each of these is divided by the diagonal from M into two (green and orange) equal areas.

### References

- S. Rabinowitz (ed),
*Problems and Solutions from the Mathematical Visitor (1877-1896)*, MathPro Press, 1996

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