What happens when $R=h_A$?(Basics of Construction)

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 $ABC$ with circumradius $R$, circumcenter $O$, and the height(altitude) of $A$ is denoted $h_A$.

To begin, we ask an essential question: when can $R=h_A$ occur? Can it happen when $\angle A$ is acute? Right? Obtuse?

Just by drawing a simple diagram, we know it cannot happen when $\angle A$ is obtuse. If $\angle A$ is obtuse, then $A,O$ are on opposite sides of $BC$ by properties of circumcenters. This means $AO$ must be larger than the closest distance from $A$ to $BC$, i.e $h_A$.

When $\angle A$ is right, $O$ becomes the midpoint of $BC$; thus the altitude and median must coincide, which makes $BAC$ an isosceles right triangle.

For the case $\angle A$ is acute, the truth is that it can always happen. We will prove this below by finding a way to construct $A$ given $BC$ and $O$. The rest of the discussion will assume that $A$ is acute. In the meantime, you are welcome to prove it any other way you like.

To kick the pure geometry off, let $K$ denote the base of the altitude of $A$. Rightaway, we will invoke the internal bisector of $\angle BAC$, which also bisects $\angle KAO$(property of circumcenter). Let the bisector intersect the circumcircle $(O)$ again at $N$, then $ON\perp BC$ because $N$ is the midpoint of minor arc $BC$. Hence $ON\parallel AK, ON=AK$, which implies $AONK$ is a parallelogram; In fact, it is a rombus because $AO=NO$. This gives us a crucial property to motivate our construction: $NO=NK$.

Construction Problem: Given a circle and on it a fixed chord $BC$. Construct $A$ such that $R=h_A$.

Solution: We know $A$ must lie on the major arc $BC$. First, construct $N$ the midpoint of minor arc $BC$. Next, draw the circle centered at $N$ with radius $ON$ intersect $BC$ at two point(because $O,N$ are on opposite sides of $BC$), denote one of the points $K$. Finally, construct perpendicular bisector of $OK$ and it intersects the major arc $BC$ at $A$, our desired construction. This ends our the construction portion.

Now we prove that $A$ indeed satisfies $R=h_A$. Since $AO=NO$, therefore $OK$ is the perpendular bisector of $AN$. Thus $AONK$ is a rombus and $AK\parallel ON$, which means $AK$ is the altitude of $A$ and $R=ON=AK=h_A$ $._{\Box}$

In conclusion, the provided construction should make it clear that $\angle A$ is acute or right is a necessary condition for $R=h_A$.

In the next note of this series, we will add the incenter of $\triangle ABC$ to the picture and derive a few awesome properties. In the mean time, ponder this:

Suppose $R=h_A$ and $I$ is the incenter of $ABC$, let $D$ be the foot of projection of $I$ on $BC$. Prove that $OD\perp ND$, where $N$ is still the midpoint of arc $BC$.