Construct a triangle ABC when angle A, length of angle bisector from A and circumradius are given.

Question

Construct a triangle ABC when angle A, length of angle bisector from A and circumradius are given.

Since angle A and circumradius are given, we can find out the length of BC. I can elaborate on the same further if you're interested... This seems like a simple problem but I can't get it. Thank you!

In a circle with a radius equal to circumradius, an angle equal to angle A constructed from any point will intersect the circle at two points, the distance between which will remain constant and equal, in fact, to side BC. This is how we can get length of side BC given angle opposite it and the circumradius.

Answer 1

I'm not sure about your first statement: I have here two circles with equal radius around the same angle $A$, but I'm quite certain that the lengths of the green line segments are different:

Answer 2

Since we have the angle and the radii we do indeed have $BC$. Here is my solution:

Let's begin with a random point $O$ of the plane to be the center of our circle $\omega = \odot (O,R)$ whose radius $R$ is given to us. Now take a random point $B$ on the circumference of $\omega$ and let $B' = BO \cap \omega \neq B$, then draw line $B'C$ ($C \in \omega$) such that $\angle BB'C = \angle A$ be the given angle we want to have on vertex $A$.

Let $M$ be the meeting of the perpendicular bissector of $BC$ with $\omega$ on the arc $\widehat{BC}$ that doesn't contain $B'$.

Now suppose we have found our solution i.e. the point $A$, let $D = AM \cap BC$. There is a famous result on geometry called the shooting lemma, which states that $MD \cdot MA = MC^2 = MB^2$.

We want $MD = x$, we have $MC = b$ and $AD = a$. All we need is to geometrically solve the equation $x \cdot (x+a) = b^2$. Although I lack a good solution, here is a bad one using a basic concept of power of a point:

• Draw line $CO' \perp MC$
• Draw circle $c_1 = \odot(C, \frac{AD}2)$
• Let $O' = c_1 \cap CO'$ and draw circle $c_2 = \odot (O',O'C)$
• Let line $MO'$ meet $c_2$ at point $D'$ closer to $M$ than to $O'$
• $D = \odot (M,MD') \cap BC$ (there are usually two solutions)
• $A = MD \cap \omega \neq M$