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Quadrilateral With Two Nonadjacent Right Angles

© Copyright 2000, Jim Loy

I received this email:

My name is Steve A. I am a student at a C.C. in Porterville Ca. We had a problem in a geometry class yesterday and I'm wondering if you might be able to help us? It is as follows: Prove: If two nonadjacent angles of a quadrilateral are right angles, then the bisectors of the other two angles either coincide or are parallel. Even the teacher had a tough time with it. In other words we could not come up with a proof. Thank you for your attention.

I responded (without the diagram), "Ah, I feel so smart. Here is what I come up with, it's fairly smooth:"

Given: quadrilateral ABCD with angles A & C right angles.

Show: bisectors of angles B & D are parallel (or coincide).

Proof:

• Draw the angle bisectors in question.
• The bisector of B intersects line AD at some point E (not necessarily between A & D by the way) [it's easy to show that this bisector and AD are not parallel]
• Angle ABE=B/2 [bisector]
• Angle AEB=90-B/2 [sum of angles in a triangle=180]
• Angle D=180-B [sum of angles in a quadrilateral=360]
• The bisector of D=90-B/2 [half of angle D]
• Therefore, the bisectors are parallel (or they may coincide if points E & D coincide) [one of the parallel theorems]

The parallel theorem is that two lines are parallel if corresponding angles on the same side of the transversal are equal. I received this response:

Unfortunately the instructor could not bring himself to use your proof. The best he could offer was a cursery glance followed by a 17 step proof. I expected we would go over it step by step pointing out where it succeeded and where it failed. I counted maybe five or six steps in yours; do you still stand by that?

I responded that my proof was a fine proof.

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