21 Oct '07 10:54

Not sure how hard/easy this is but I found a neat solution.

Consider two touching circles. Let their common tangent be the line T. Take any point on T, lets call it P.

Now take a line from P through the centre of the first circle so that this line cuts the circle at points A & B.

Similarly for the second circle to get points C & D.

Prove that triangle PAC is similar to triangle PDB.

Consider two touching circles. Let their common tangent be the line T. Take any point on T, lets call it P.

Now take a line from P through the centre of the first circle so that this line cuts the circle at points A & B.

Similarly for the second circle to get points C & D.

Prove that triangle PAC is similar to triangle PDB.