### 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.

 What if applet does not run?

Explanation

The applet may suggest the following statement:

 Let point P lie on a side AB of ΔABC. Circle C(A) is the circumcircle of ΔAPC, with circumcenter at O. Circle C(B) is the circumcircle of ΔBPC, with circumcenter at O'. AO' crosses C(A) in M. CM crosses C(B) in D. Show that O lies on AC iff O' lies on BC. If O lies on AC then CM = DM.

### 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.

 What if applet does not run?

### Proof

If O lies on AC, AC is the diameter of C(A) and hence any inscribed angle subtended by AC is right. In particular, ∠APC is right. But then its supplementary angle BPC is also right and is inscribed into C(B). The chord it is subtended by must be a diameter in C(B). Therefore, O' lies on BC. The argument is obviously reversible.

If O lies on AC, then ∠AMC is also right (inscribed in C(A)) and so is ∠CDB (inscribed in C(B)). ∠CMO' is right as supplementary to ∠AMC. We conclude that two triangles CDB and CMO' that share an angle at C have angles CDB and CMO' also equal. They are thus similar. From here, MO'||BD. Since BO' = CO' we also have BM = DM.