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 $ABC$.


 * ''If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle and the other fall on it, and if further 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 straight line which falls on the circle, the straight line which falls on it will touch the circle.

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 $ABC$.