Construction of Tangent from Point to Circle/Proof 2
Theorem
From a given point outside a given circle, it is possible to draw a tangent to that circle.
In the words of Euclid:
- From a given point to draw a straight line touching a given circle.
(The Elements: Book $\text{III}$: Proposition $17$)
Proof
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.
$\blacksquare$
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.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions