Given a segment AC and a random point B on it, construct the semicircles of diameters AB and AC and let F be midpoint of AB. A circle of center E is internally tangent to the semicircle (AC), externally tangent to semicircle (AB) and tangent to the tangent at B to semicircle (AB). Let the circle E touch the semicircle AB at P, and PF cut the tangent at B to semicircle (AB) at G. Prove that PG=BC.
Proof
Note that the circle E is needed to get an idea of the position of P only: inversion of pole C and power CB.CA sends the semicircle (AC) to the line GB, while the semicircle (AB) and circle E remain unchanged, thus PC is tangent at P to semicircle (AB). As FP=FB ( 1 ) we infer that triangles FBG and FPC are congruent, FG=FC ( 2 ) and with (1) we get the desired PG=BC.
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