Prerequisite: Triangles - Circumcenter
In this series of notes, I will use pure geometry to deduce a few properties in the configuration where a triangle's circumradius is equal to one of its heights. Our discussion will pertain to the main triangle with circumradius , circumcenter , and the height(altitude) of is denoted .
To begin, we ask an essential question: when can occur? Can it happen when is acute? Right? Obtuse?
Just by drawing a simple diagram, we know it cannot happen when is obtuse. If is obtuse, then are on opposite sides of by properties of circumcenters. This means must be larger than the closest distance from to , i.e .
When is right, becomes the midpoint of ; thus the altitude and median must coincide, which makes an isosceles right triangle.
For the case is acute, the truth is that it can always happen. We will prove this below by finding a way to construct given and . The rest of the discussion will assume that is acute. In the meantime, you are welcome to prove it any other way you like.
To kick the pure geometry off, let denote the base of the altitude of . Rightaway, we will invoke the internal bisector of , which also bisects (property of circumcenter). Let the bisector intersect the circumcircle again at , then because is the midpoint of minor arc . Hence , which implies is a parallelogram; In fact, it is a rombus because . This gives us a crucial property to motivate our construction: .
Construction Problem: Given a circle and on it a fixed chord . Construct such that .
Solution: We know must lie on the major arc . First, construct the midpoint of minor arc . Next, draw the circle centered at with radius intersect at two point(because are on opposite sides of ), denote one of the points . Finally, construct perpendicular bisector of and it intersects the major arc at , our desired construction. This ends our the construction portion.
Now we prove that indeed satisfies . Since , therefore is the perpendular bisector of . Thus is a rombus and , which means is the altitude of and
In conclusion, the provided construction should make it clear that is acute or right is a necessary condition for .
In the next note of this series, we will add the incenter of to the picture and derive a few awesome properties. In the mean time, ponder this:
Suppose and is the incenter of , let be the foot of projection of on . Prove that , where is still the midpoint of arc .