Angle Bisectors In Rectangle

What is this about?
A Mathematical Droodle

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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to suggest the following problem [Prasolov, p. 12]:

ABCD is a rectangle; M and N are the midpoints of sides AD and BC, respectively. Let P lie on CD, Q be the intersection of MP and AC. Prove that MN is the bisector of ∠PNQ.

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Since MN||CD triangles CPQ and OMQ are similar, implying the proportion

MP/MQ = CO/OQ.

Let O be the midpoint of AC (i.e., the center of the rectangle) and K on NQ be such that KO||BC. Then triangles CNQ and OKQ are similar, implying the proportion

CO/OQ = KN/KQ.

By the transitivity of equality, we have a proportion in triangles NPQ and KMQ

MP/MQ = KN/NQ.

which implies KM||NP. We are almost done.

ΔKMN is isosceles (because MO = NO and KO⊥MN) so that ∠KMN = ∠KNM. But, since KM||NP, the vertical angles KMN and MNP are also equal, giving the required ∠QNM = ∠MNP.

References

  1. V. V. Prasolov, Problems in Planimetry I, Nauka, 1986 (in Russian)

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  • The Nature of Pi
  • Parallelogram and Similar Triangles
  • Two Triples of Similar Triangles
  • Three Similar Triangles
  • Similar Triangles on Sides and Diagonals of a Quadrilateral
  • Point on Bisector in Right Angle

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    Copyright © 1996-2018 Alexander Bogomolny

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