The figure below shows an arbelos ABC (AB, BC, and AC are semicircles of centers O_{1}, O_{2}, and O).
The incircle I of the arbelos is tangent to semicircles AC, AB, and BC at T,
T_{1}, and T_{2}, 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

All Problems

Arbelos

Triangle

Circle

Semicircle

Tangent Circles

Incircle

Angle Bisector

Angle

Perpendicular lines

Midpoint

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