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# Surprising Perpendicularity - Problem 1

In $$\triangle ABC$$ with $$\angle ABC=90^\circ$$, let $$D$$ and $$E$$ be the intersection of the angle bisectors of $$\angle CAB$$ and $$\angle BCA$$ with $$BC$$ and $$AB$$, respectively. If $$I$$ is the intersection of $$AD$$ and $$CE$$, and $$O$$ is the circumcenter of $$\triangle BED$$, show that $$OI\perp CA$$.

Note by Cody Johnson
3 years, 9 months ago

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Let $$D',J,E'$$ be the projections of $$D,I,E$$ on $$AC$$. Since $$O$$ is just the midpoint of $$ED$$, we only have to prove $$JD'=JE'$$. By the bisector and Pythagoras' theorems we have: $CD'=CD\cos C=CD\sin A=\frac{ab}{b+c}\sin A=\frac{a^2}{b+c}=\frac{a^2(b-c)}{b^2-c^2}=b-c,$ and $$AE'=b-a$$. Since $$AJ=p-a$$ and $$CJ=p-c$$, the claim follows.

- 3 years, 9 months ago