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this is question 42 in the red book of mathematical problems by k. s. williams and k. hardy.

let abcd be a convex quadrilateral. let p be the point outside abcd such that $|ap| = |pb|$ and $\angle apb = 90^\circ.$ the points $q, r, s$ are similarly defined. prove that the lines $pr$ and $qs$ are equal length and perpendicular.

you can make a physical model by cutting four isoscles right triangles of various lengths from a rectangular card. spread them out on a table with the right angles in the exterior. the result sated in the problem can be seen to hold.

the solution in the book use complex numbers and i can solve using the cosine rule. i am wondering if there is a more "geometric" solution with as little as computation as possible.

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Did you mean that $pr$ and $qs$ are of equal length and perpendicular? – Isaac Apr 22 '11 at 4:19
isaac, yes. i have edited the post. – abel Apr 27 '11 at 14:35
up vote 5 down vote accepted

I think that using complex numbers is the most geometric thing to try, because multiplication by $i$ gives you a rotation by $90^\circ$ for free, i.e. without choosing a basis. In fact, treating all occurring points as complex numbers one immediately obtains $s-q=i(r-p)$ without any squaring or trigonometry coming into the game.

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