Constructing a Tangent

I was recently asked for an elegant proof of the following problem. It’s based on a construction challenge from Euclidea.

Given: Circles A, B, and C, such that point C is on circle A, point B is on circles A and C, point E is on circles B and C, and point D is on all three circles.

Prove: $\overleftrightarrow{BE}$ is tangent to circle A.

In Euclidea, the challenge is to construct the tangent on circle A through point B given only those two objects. The most efficient solution is:

1. Draw a circle with center C (arbitrarily placed) on A, with B on that circle.
2. Draw a circle with center B, such that the intersection of circles A and C is on B.
3. Draw a line through the other intersection of circles B and C. This will be tangent to A at B.

The challenge I was given was to create as elegant a proof as possible. I have a proof; whether it is sufficiently elegant is a matter of opinion.

Here is a closeup of the diagram. I have drawn three triangles. The goal is to prove that $m\angle ABE = 90^\circ$.

Each of these triangles is isosceles; $\Delta BCD$ and $\Delta BCE$ each have two legs that are radii of circle C, while $\Delta BAD$ has two legs that are radii of circle A. Since they are radii of circle B, $\overline{BE}\cong\overline{BD}$. By SSS, $\Delta BCD \cong \Delta BCE$.

We can now isolate the concave pentagon and examine its interior angles.

Let $\alpha = m\angle CBD = m\angle CDB = m\angle CBE = m\angle CEB$, $\beta = m\angle BCD = m\angle BCE$, $\delta = m\angle BAD$, and $\gamma = m\angle ABD = m\angle ADB$.

By the Triangle Sum Theorem, we know that $2\alpha + \beta = 2\gamma + \delta = 180^\circ$.

Inscribed angles have half the measure of central angles, and inscribed $\angle BCD$ corresponds to the reflex of central $\angle BAD$, so $2\beta = 360^\circ – \delta$; thus, $\delta = 360^\circ – 2\beta$. By substitution, $2\gamma + 360^\circ – 2\beta = 180^\circ$. Simplify this to $\beta – \gamma = 90^\circ$.

Since we know $\beta = 180^\circ – 2\alpha$, substitute again to $180^\circ – 2\alpha – \gamma = 90^\circ$, that is, $2\alpha + \gamma = 90^\circ$.

We can see that $m\angle ABE = 2\alpha + \gamma = 90^\circ$. Thus $\overline{EB} \perp \overline{AB}$, QED.