Tangent Secant Theorem

From ProofWiki
Jump to navigation Jump to search

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 1

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.

$\Box$


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

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

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.

$\blacksquare$


Proof 2

Let the circle $\CC$ be embedded in a Cartesian plane with its center located at the origin.

Let $\CC$ have radius $r$.

From Equation of Circle center Origin, $\CC$ can be described as:

$(1): \quad x^2 + y^2 = r^2$


From Equation of Straight Line in Plane: Parametric Form, let the straight line $DCA$ be described using the parametric equations:

$\begin {cases} x = x_0 + \rho \cos \psi \\ y = y_0 + \rho \sin \psi \end {cases}$

where $D = \tuple {x_0, y_0}$.


Substituting these for $x$ and $y$ into equation $(1)$:

\(\ds \paren {x_0 + \rho \cos \psi}^2 + \paren {y_0 + \rho \sin \psi}^2\) \(=\) \(\ds r^2\)
\(\ds \leadsto \ \ \) \(\ds x_0^2 + 2 x_0 \rho \cos \psi + \rho^2 \cos^2 \psi + y_0^2 + 2 y_0 \rho \sin \psi + \rho^2 \sin^2 \psi\) \(=\) \(\ds r^2\)
\(\ds \leadsto \ \ \) \(\ds \rho^2 \paren {\cos^2 \psi + \sin^2 \psi} + 2 \rho \paren {x_0 \cos \psi + y_0 \sin \psi} + x_0^2 + y_0^2\) \(=\) \(\ds r^2\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \rho^2 + 2 \rho \paren {x_0 \cos \psi + y_0 \sin \psi} + x_0^2 + y_0^2 - r^2\) \(=\) \(\ds 0\) Sum of Squares of Sine and Cosine

We see that $(2)$ is a quadratic equation in $\rho$.

Let $\rho_1$ and $\rho_2$ be the solutions of $(2)$.

Each of these corresponds to one of the points at which $DCA$ intersects $\CC$, where:

$\rho_1 = DA$
$\rho_2 = DC$

From Product of Roots of Quadratic Equation:

$\rho_1 \rho_2 = x_0^2 + y_0^2 - r^2$

From Length of Tangent from Point to Circle center Origin:

$DB = x_0^2 + y_0^2 - r^2$

Hence we have:

$DA \cdot DC = DB^2$

$\blacksquare$