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Explanation

The applet may suggest the following statement:

 Let point P lie on a side AB of ΔABC. Circle C(A) passes through A and P and is tangent to AC, circle C(B) passes through B and P and is tangent to BC. Q is the second point of intersection of C(A) and C(B). Show that, regardless of the position of P on AB, line PQ passes through a fixed point X.

### 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.

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### Proof

First note that in C(A)

 ∠AQP = ∠CAP.

Similarly, in C(B),

 ∠BQP = ∠CBP.

If so, then

 ∠AQB + ∠ACB = 180°,

which says that quadrilateral ACBQ is cyclic so that Q lies on the circumcircle of ΔABC.

Let X be the point of intersection of PQ with the circumcircle. Inscribed angles AQX and BAC are equal. Therefore, the arcs AX and BC they subtend are also equal. It follows that CX is parallel to AB. Hence, X is independent of P.