# Angle Bisectors In Rectangle

## What is this about?

A Mathematical Droodle

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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

### References

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Copyright © 1996-2018 Alexander Bogomolny