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Given a rectangle WXYZ, let R be a point on its circumscribed circle. Show that, out of the orthogonal projections of R onto WX, XY, YZ, and ZW; one out of these 4 points is the orthocenter of the triangle created by the other three.

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    Have you tried saying $W,X,Y,Z$ are the points $(\pm a,\pm b)$ and $R=(r\cos\theta,r\sin\theta)$ for $r=\sqrt{a^2+b^2}$? Then the projections are $R_{1,2}=(\pm a,r\sin\theta)$ and $R_{3,4}=(r\cos\theta,\pm b)$ and it's fairly straightforward to check that each of the three pairs of lines $\{R_aR_b,R_cR_d\}$ are perpenticular. Thus each of the $R_i$ lies on the altitudes of the triangle formed by the other three points, proving the result. Remember that perpendicularity means the slopes are negative reciprocals, or that their products are $-1$.2012-03-19

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