Geometric proof that $\tan \alpha \geq \alpha$ for any acute angle $\alpha$

Question

For any acute angle $\alpha$, then $\tan \alpha \geq \alpha$. Is it possible to prove this geometrically?

My work so far:

A rigorous proof requires defining a radian measure of angles: this can be done in one of two ways:

1. Arclength. E.g. define a semicircle to be $\pi$, and the measure of angle to be the portion of the semicircle it spans.
2. Area. E.g. define the area of a circle to be $\pi$, and the measure of angle to be its area compared to the area of a semicircle.

Using arclength seems simpler (although area may end up being more rigorous), so I'll start with that.

There is a similar identity $\sin \alpha \leq \alpha$. This follows directly from the triangle inequality:

$\overset\frown {BD} \geq BD \geq BC$.

I tried to do something with the tangent segment (not show on the diagram, but the segment from $D$ parallel to $BC$ until intersecting line $AB$), but was unable to prove this.