geometric-constructiongeometry
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.
If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$.
$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.
See the drawing below:
It is possible, provided that the point and at least part of the circle are on the same side of the line and the circle does not separate the point from the line. If the point and the circle are on opposite sides of the line, or if the point is inside the circle and the line wholly outside, then it is impossible (as it is with three nested circles in the original version).
Wikipedia describes this as Special case 6 of Apollonius' problem with up to 4 solutions
Best Answer
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
Now, the proof :
You have that :
This gives you that, for the triangle APP',
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