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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.

Jack D'Aurizio - 3 years, 9 months ago

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