The figure below shows an arbelos ABC (AB, BC, and AC are semicircles of centers O1, O2, and O). The incircle I of the arbelos is tangent to semicircles AC, AB, and BC at T, T1, and T2, respectively. The bisector of the angle ATC cuts the incircle at P and IP extended cuts AC at H. Prove that (1) IH is perpendicular to AC; (2) P is the midpoint of IH.
Click on the figure below.
Geometry Problems
1291-1300
Visual Index
Open Problems
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Arbelos
Triangle
Circle
Semicircle
Tangent Circles
Incircle
Angle Bisector
Angle
Perpendicular lines
Midpoint
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