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Given $\triangle ABC$. On the side AB externally is constructed square ABPQ. On the side AC internally is constructed square ACMN. AH is the altitude. If $O_1$ and$O_2$ are the centers of the two squares, prove that $O_1, O_2 $ and $H$ are collinear.

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Is $H$ the intersection of the altitudes or the intersection of the altitude with the opposite side? –  Phira Jun 4 '12 at 20:23
    
From the question, $AH$ is an altitude, so $H$ lies on $BC$. –  Théophile Jun 4 '12 at 20:44
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Let $A'$ be the intersection of the altitude through $A$ with $BC$ (the point called $H$ by the OP).

Let $X$ be the point on $BC$ such that $AA'X$ forms an isosceles rectangular triangle. (The vector $BC$ points in the same direction as the vector $A'X$.)

Then, a clockwise rotation by $45^\circ$ around $A$ followed by a scaling by $\frac1 {\sqrt 2}$ moves the points $X,C,B$ to the points $A',O_2,O_1$.

Clearly, a line remains a line after rotating and scaling.

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Nicely done. A couple of comments, though: first, why rename (and redefine) $H$? More importantly, there are two possibilities for $X$, one on either side of $H$. You can avoid this problem by defining $X$ to lie on $CH$. –  Théophile Jun 4 '12 at 21:00
    
I redefined it because $H$ always denotes the orthocenter. It is like starting a question with "Let $i$ denote the square root of $-2$.". As to your other remark, I considered defining $X$ by a rotation, but it is a convention that the points of a triangle are usually given in counter-clockwise order which fixes the position of $X$. I don't understand your suggestion, because $X$ does not necessarily lie on the segment $CH$ or on the half-line $CH$. And the line $CH$ is no different from the line $BC$ that I used. But I will add a clarification. –  Phira Jun 4 '12 at 21:28
    
Beautiful solution indeed. Do you have any idea how can I avoid using the rotation? –  Adam Jun 4 '12 at 23:29
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