Construction of Tangent from Point to Circle/Proof 2

Proof

 * Euclid-III-17a.png

Let $BCD$ with center $A$ be the circle, and let $E$ be the exterior point from which a tangent is to be drawn.

Bisect $AE$ at $F$.

Then draw a circle $AEG$ whose center is $F$ and whose radius is $AF$.

The point $G$ is where $AEG$ intersects $BCD$.

The line $EG$ is the required tangent.

Proof of Construction
By the method of construction, $AE$ is the diameter of $AEG$.

By Thales' Theorem $\angle AGE$ is a right angle.

But $AG$ is a radius of $BCD$.

The result follows from Radius at Right Angle to Tangent.

Historical Note
An easier solution is of course possible as soon as we know $(\text{III}$. $31)$ that the angle in a semicircle is a right angle; for we have only to describe a circle on $AE$ as diameter, and this circle cuts the given circle in the two points of contact.
 * -- : Footnote to Book $\text{III}$ Proposition $19$