# 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

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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Angles BAP and CAQ are vertical,vertical,alternate,exterior and hence equal. Triangles BAP and CAQ are isosceles,right,equal,isosceles with equal base angles at A. The other pair of the base,alternate,base 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