Secant-Tangent TheoremDate: 03/21/2002 at 17:59:31 From: Jon Subject: Secant-Tangent Theorem Hi, I'm trying to prove the secant-tangent theorem, but I'm stuck. I'm pretty sure it involves showing two triangles are similar, and then relating the proper ratios. The secant-tangent theorem states the following: If a secant PA and tangent PC meet a circle at the respective points A, B, and C(point of contact), then (PC)^2=(PA)*(PB). ["*" denotes regular multiplication, and PC, PA, PB are all segment lengths]. Also, on another note, how can we show that there is a line tangent to a circle C through any point P outside of the circle? Any help would be great. - Jon Date: 03/22/2002 at 17:38:00 From: Doctor Mitteldorf Subject: Re: Secant-Tangent Theorem Dear Jon, The trick is to show that triangle PAC is similar to triangle PCB. These triangles already have one angle in common. Can you prove that angle CAB equals angle PCB? Hint: The arc BC is twice the inscribed angle CAB. Draw the isosceles triangle AOB, where O is the center of the circle, and angle PCB becomes an exterior angle, minus a right angle. I hope this is enough to get you going, but not enough so you can't still have fun with this proof. This theorem is a special case of Steiner's Theorem, where one of the secants is a tangent (i.e., the two points defining the secant become a single point). See: Steiner's Theorem - June Jones http://jwilson.coe.uga.edu/emt669/Student.Folders/Jones.June/steiner/steinerthm.html - Doctors Mitteldorf and Ian, The Math Forum http://mathforum.org/dr.math/ |
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