Outline Mathematics
Geometry

Two Touching Circles

Consider the following problem:

 

Two circles with centers P and Q touch at point A. A line through A meets the first circle again at B and the second at C. Show that BP || CQ.



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Solution

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Copyright © 1996-2012 Alexander Bogomolny

 

Two circles with centers P and Q touch at point A. A line through A meets the first circle again at B and the second at C. Show that BP || CQ.



This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


(In the text below, some words are omitted. These have been underlined. Click just above the line. See what happens.)

Angles BAP and CAQ are and hence equal. Triangles BAP and CAQ are with equal base angles at A. The other pair of the angles are also equal. I.e., ∠ABP = ∠ ACQ. These two angles are internal to the lines BP and CQ and transversal BC. Since the internal angles are equal, the lines are parallel: PB || QC.

(The terms you met: Vertical angles, Alternate angles, Transversal, Isosceles triangle)

References

  1. V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986, in Russian

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Copyright © 1996-2012 Alexander Bogomolny

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