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/
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