Equivalence of Definitions of Cotangent of Angle

Theorem
Let $\theta$ be an angle.

Definition from Triangle implies Definition from Circle
Let $\cot \theta$ be defined as $\dfrac {\text{Adjacent}} {\text{Opposite}}$ in a right triangle.

Consider the triangle $\triangle OAB$.

By construction, $\angle OAB$ is a right angle.

From Parallelism implies Equal Alternate Interior Angles, $\angle OBA = \theta$.

Thus:

That is:
 * $\cot \theta = AB$

Definition from Circle implies Definition from Triangle
Let $\cot \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$.

Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above.

From Parallelism implies Equal Alternate Interior Angles, $\angle OBA = \theta$.

We have that:
 * $\angle CAB = \angle OBA = \theta$
 * $\angle ABC = \angle OAB$ which is a right angle

Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar.

By definition of similarity:

That is:
 * $\dfrac {\text{Adjacent} } {\text{Opposite} } = \cot \theta$