PQ is actually a diameter of the circle so P is the midpoint of the arc AB.
Assume C is on the opposite side of AB from P(*). The incenter is on the bisector of angle ACB so angles ACI and ICB are equal and the ray CI passes through P.
Now turning our attention to triangle AIB, we have the perpendicular bisector of AB already (PQ), when we construct the other two perpendicular bisectors we find H, which is the circumcenter of triangle AIB. So H is on ray CI. But H is also on PQ. Therefore CI and PQ cross at H and H=P.
(*)If C is on the same side of AB as P, interchange P and Q in the above.
Edited to replace the wrong term excenter with circumcenter.
(The intersection of perpendicular bisectors of the sides of a triangle.)
Edited on January 16, 2018, 1:32 pm
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Posted by Jer
on 2018-01-15 15:10:09 |
Solution
