Converse of Tangent Secant Theorem

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

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

Let $DB$ intersect the circle at $B$ such that $DB^2 = AD \cdot DC$.

Then $DB$ is tangent to the circle $ABC$.

Proof

 * Euclid-III-37.png

Let $DE$ be drawn tangent to the circle $ABC$.

Let $F$ be the center of $ABC$ and join $FB, FD, FE$.

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

We have that $DE$ is tangent to the circle $ABC$ and $DA$ cuts it.

So from the Tangent Secant Theorem $AD \cdot DC = DE^2$.

But we also have by hypothesis that $AD \cdot DC = DB^2$.

So $DE^2 = DB^2$ and so $DE = DB$.

Also $FE = FB$ and so $DE, EF = DB, DF$ and $DF$ is common.

So from Triangle Side-Side-Side Equality $\angle DEF = \angle DBF$.

But $\angle DEF$ is a right angle and so also is $\angle DBF$.

So from Line at Right Angles to Diameter of Circle $DB$ is tangent to the circle $ABC$.