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:
- Draw a circle with center C (arbitrarily placed) on A, with B on that circle.
- Draw a circle with center B, such that the intersection of circles A and C is on B.
- 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.