Tangent Secant Theorem

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 $DB^2 = AD \cdot DC$.

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.


 * Euclid-III-36a.png

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.

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

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


 * Euclid-III-36b.png

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.