Paul recently asked for a synthetic proof of a neat construction ...

> Given triangle ABC, it is easy to construct the two circles,

>each passing through A, and tangent to BC at one of the vertices B

>and C. How does one construct the circle tangent to BC, and to each of these

>two circles externally?

>Let the bisector of angle A intersect BC at P. Construct the

>perpendicular to BC at P to intersect the median AD at Q.

>The cirlce with PQ as diameter is tangent BC and to each of the two

>circles above.

He mentioned a construction by inversion and I wondered whether this would

not also prove his direct one. So I start with what is going to be an

inverse figure ....

Consider a triangle ABC with internal bisector at A meeting the

circumcircle S at P and with tangents at B,C meeting at K. A circle S1

touches these tangents and the circle S at P. KA meets the circles S,S1

in corresponding points A,A1 and Q,Q1; KP meets S1 again at P1. Note

angles QAP=Q1A1P1=Q1PP1 so that the circle Q1PA touches KP at P, ie is

orthogonal to the circle S.

Inverting w.r.t. A proves Paul's construction (with angles B,C

interchanged). This invokes the fact that the adjoint circles through A

(touching BC at B, C respectively) meet on the median from A (in the above

case this intersection is the inverse of K)

Dick Tahta