# Construction of Tangent from Point to Circle/Proof 1

## 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.

## Proof Let $A$ be the given point and let $BCD$ be the given circle.

It is required that a straight line be drawn from $A$ to $BCD$.

Join $AE$ and draw the circle $AFG$ with center $E$ and radius $AE$.

Join $EF$ and let $B$ be the point at which $EF$ joins the circle $BCD$.

Join $AB$.

Then $AB$ is the required tangent to $BCD$.

### Proof of Construction

Since $E$ is the center of the circles $BCD$ and $AFG$, $EA = EF$ and $ED = EB$.

Therefore the two sides $AE, EB$ equal the two sides $FE, ED$, and they contain a common angle at $E$.

So by Triangle Side-Angle-Side Equality, $\triangle DEF = \triangle BEA$.

Therefore $\angle EDF = \angle EBA$.

But $\angle EDF$ is a right angle, so $\angle EBA$ is also a right angle.

But $EB$ is a radius of circle $BCD$.

It follows from the porism to Line at Right Angles to Diameter of Circle that $AB$ is tangent to $BCD$.

$\blacksquare$

## Historical Note

This theorem is Proposition $17$ of Book $\text{III}$ of Euclid's The Elements.
Note that Euclid fails to specify that the point needs to be outside the circle, and also completely ignores the possibility that the point be on the circumference. The latter could of course be considered to be covered by the porism to Book $\text{III}$ Proposition $16$: Line at Right Angles to Diameter of Circle.

It is also not pointed out that there are in fact two tangent to the circle from such an exterior point, the second of which can be found by extending $FD$ past $D$ to intersect circle $AFG$ on the other side.