Constructing a Tangent

Tangent to a CircleI 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.

In addition to the circles, 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, \(\Delta BCD \cong \Delta BCE\).

Tangent to a Circle 2We can now isolate the concave pentagon and examine its interior angles. I’ve labeled all congruent angles with the same Greek letters in the diagram, and will use them to complete the proof.

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 \(\angle DCB\) corresponds to the reflex of central \(\angle DAB\), so \(2\beta = 360^\circ Р\delta\); thus, \(\delta = 360^\circ Р2\beta\). By substitution, \(2\gamma + 360^\circ Р2\beta = 180^\circ\). Simplify this to \(\beta Р\gamma = 90^\circ\).

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.

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