Given a circle, its diameter and an external point, use a straightedge to draw a line through the point and perpendicular to the diameter Some time back I saw the following problem which originated in Russia: > You are given a circle, its diameter and an external point not on the diameter (A, B and P in the diagram below). Using only a straightedge, construct a line through the point that is perpendicular to the diameter. Prove that the constructed line is indeed perpendicular. Consider only the case where the perpendicular line meets the diametral line at a point Q that is outside the circle. !enter image description here

Call A' the intersection of the circle with AP Call B' the intersection of the circle with BP Call P' the intersection of A'B and AB'

Then PP' is perpendicular to AB

!enter image description here

Now, the proof :

You have that :

* The triangle AA'B is a right triangle because it's a triangle inscribed in a circle with one of its side as a diameter.
* The triangle AB'B is a right triangle because it's a triangle inscribed in a circle with one of its side as a diameter.



This gives you that, for the triangle APP',

* PB' is the altitude from the vertex P
* P'A' is the altitude from the vertex P'



So B, the intersection of PB' and P'A', is the orthocentre.

It follow that AB is the altitude from the vertex A, hence AB and PP' are perpendicular