Symmetry Proof

Date: 09/27/2001 at 01:49:00
From: Jane Roos
Subject: Geometry

Given an angle with vertex O and a point P inside the angle, drop 
perpendiculars PA, PB to the two sides of the angle, draw AB, and drop 
perpendiculars OC, PD to line AB. Then show that AC = BD.

I've tried to do this using parallelograms where I extend DP to get a 
point E and make a parallelogram AEBO, and we make AE parallel to OB. 
I can't figure a way to justify that AO is parallel to EB. 

Help please, thanks,

Jane

Date: 09/27/2001 at 06:07:21
From: Doctor Floor
Subject: Re: Geometry

Hi, Jane, thanks for writing.

Let us find a point Q making BPAQ into a parallelogram, and a point L 
making OBLA into a parallelogram:

        A-------------L
       /|\`.         /
      / | C `.      /
     /  Q  \  `.P  /
    /    `. D  |  /
   /       `.\ | /
  O------------B

If we take a closer look at this, we see that

  * AQ is parallel to BP and thus perpendicular to OB,
  * BQ is parallel to AP and thus perpendicular to OA,

and thus Q is the orthocenter of triangle OAB, and line OC, being the 
third altitude of this triangle, passes through Q.

In a similar way way the line PD passes through L, being the third 
altitude apart from PB and AP in triangle ABL.

That AC = BD now follows from symmetry of the figure, in particular 
from triangles OBA and LAB being congruent.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/