# Straightedge-only construction of a perpendicular

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.

• 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. – Chris Sep 1 '13 at 3:19
• 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. – imranfat Sep 1 '13 at 3:34
• I can't sketch it out right now, but some right triangles should show up. Maybe use similarity? – dfeuer Sep 1 '13 at 4:12
• @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. – Ted Shifrin Sep 1 '13 at 4:19
• @TedShifrin Per my comment below (in case it didn't ping you), thanks for the tip! – Chris Sep 1 '13 at 4:55

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. 