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There is a circle in the plane with a drawn diameter. Given a point inside the circle (not on the diameter or the circle), draw the perpendicular from the point to the diameter using only a straightedge.

Source: This question. It is about to be closed for containing too many problems in one question. I'm posting each problem separately.

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  • $\begingroup$ A construction for this is as follows: draw lines from the endpoints $E_1$ and $E_2$ of the circle through the point (call it $A$), meeting the circle in $B$ and $C$; now extend $E_2C$ and $E_1B$ until they meet in $D$, and draw $DA$ for the perpendicular. I have no proof of this, however. $\endgroup$ – Chris Sep 1 '13 at 3:19
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    $\begingroup$ The proof can be found in the field of Projective Geometry. I don't remember the proof either, but I vividly remember this construction from my projective geometry classes. $\endgroup$ – imranfat Sep 1 '13 at 3:34
  • $\begingroup$ I can't sketch it out right now, but some right triangles should show up. Maybe use similarity? $\endgroup$ – dfeuer Sep 1 '13 at 4:12
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    $\begingroup$ @user1296727: You're right on the money. The proof is simple: The altitudes of $\triangle DE_1E_2$ are concurrent; $E_1B$ and $E_2B$ are altitudes meeting at $A$. Therefore, $DA$ is perpendicular to $E_1E_2$. Please post it all as a solution. $\endgroup$ – Ted Shifrin Sep 1 '13 at 4:19
  • $\begingroup$ @TedShifrin Per my comment below (in case it didn't ping you), thanks for the tip! $\endgroup$ – Chris Sep 1 '13 at 4:55
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Solution: Taking it as known that the perpendicular bisectors of a triangle are concurrent, and that this implies the concurrence of the altitudes of a triangle, our construction and proof are simple.

Draw lines from the endpoints $E_1$ and $E_2$ of the circle through the point (call it $A$), meeting the circle in $B$ and $C$; now extend $E_1C$ and $E_2B$ until they meet in $D$. Now, by Thales' theorem, $E_1B$ and $E_2C$ are altitudes of $\triangle E_1DE_2$, meeting in the point $A$. Thus, the third altitude of this triangle, dropped from $D$, also passes through $A$; that is, $DA$ is the third altitude of the triangle. As such, it is perpendicular to the diameter of the circle. enter image description here

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  • $\begingroup$ And thanks to @TedShifrin for pointing me in the right direction! $\endgroup$ – Chris Sep 1 '13 at 4:54
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    $\begingroup$ I've fixed a couple of typos and added an image to help with the understanding; feel free to remove it if you don't like it. $\endgroup$ – ShreevatsaR Sep 2 '13 at 5:06

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