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### 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.

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/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Symmetry/Tessellations

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