# Tangent Secant Theorem/Proof 1

## Theorem

Let $D$ be a point outside a circle $ABC$.

Let $DB$ be tangent to the circle $ABC$.

Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.

Then $DB^2 = AD \cdot DC$.

In the words of Euclid:

*If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the tangent.*

(*The Elements*: Book $\text{III}$: Proposition $36$)

## Proof

Let $DA$ pass through the center $F$ of circle $ABC$.

Join $FB$.

From Radius at Right Angle to Tangent, $\angle FBD$ is a right angle.

We have that $F$ bisects $AC$ and that $CD$ is added to it.

So we can apply Square of Sum less Square and see that:

- $AD \cdot DC + FC^2 = FD^2$

But $FC = FB$ and so:

- $AD \cdot DC + FB^2 = FD^2$

But from Pythagoras's Theorem we have that $FD^2 = FB^2 + DB^2$ and so:

- $AD \cdot DC + FB^2 = FB^2 + DB^2$

from which it follows that:

- $AD \cdot DC = DB^2$

which is what we wanted to show.

$\Box$

Now let $DA$ be such that it does not pass through the center $E$ of circle $ABC$.

Draw $EF$ perpendicular to $DA$ and draw $EB, EC, ED$.

From Radius at Right Angle to Tangent, $\angle EBD$ is a right angle.

From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AC$.

So $AF = FC$.

So we can apply Square of Sum less Square and see that:

- $AD \cdot DC + FC^2 = FD^2$

Let $FE^2$ be added to each:

- $AD \cdot DC + FC^2 + FE^2 = FD^2 + FE^2$

Now $\angle DFE$ is a right angle and so by Pythagoras's Theorem we have:

- $FD^2 + FE^2 = ED^2$
- $FC^2 + FE^2 = EC^2$

This gives us:

- $AD \cdot DC + EC^2 = ED^2$

But $EC = EB$ as both are radii of the circle $ABC$.

Next note that $\angle EBD$ is a right angle and so by Pythagoras's Theorem we have:

- $ED^2 = EB^2 + DB^2$

which gives us:

- $AD \cdot DC + EB^2 = EB^2 + DB^2$

from which it follows that:

- $AD \cdot DC = DB^2$

which is what we wanted to show.

$\blacksquare$

## Historical Note

This proof is Proposition $36$ of Book $\text{III}$ of Euclid's *The Elements*.

It is the converse of Proposition $37$: Converse of Tangent Secant Theorem.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions