Date: 06/09/2002 at 15:33:36 From: Elfspirit Subject: Geometry- urgent Hello. Here is my math problem: There are two chords inscribed in a circle, AB and CD. They intersect at point E. Prove that AE * EB = CE * ED. First, I drew chords AD and CB to create two triangles. The vertical angles AED and CEB are congruent, and so are angles AEC and DEB. But I am lost from there. I know I should prove that the two triangles formed are similar triangles, but I'm still not entirely certain how much that will aid me in this proof. Is it possible the proof was phrased wrong and it means that AE * EB is PROPORTIONAL to CE * ED? Please, please, please answer as soon as you possibly can; this is very important.
Date: 06/10/2002 at 03:52:40 From: Doctor Floor Subject: Re: Geometry- urgent Hi, Thanks for your question. Consider the following figure: Indeed angles <AEC and <BED are congruent vertical angles. Also <CAB and <CDB are congruent, because they are inscribed angles on the same arc. In a similar fashion <ACD and <ABD are congruent. This shows us that triangles BED and CEA are similar (note that B corresponds to C and D to A). From that: BE/CE = ED/EA BE*EA = CE*ED If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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