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Secant-Tangent Theorem

Date: 03/21/2002 at 17:59:31
From: Jon
Subject: Secant-Tangent Theorem


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   

- Doctors Mitteldorf and Ian, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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