A circle can have any number of secants from a point outside the circle. The secants drawn to a circle from the same point outside the circle are called intersecting secants. There are a few theorems related to secants of a circle. Let us state and prove some of the important theorems on intersecting secants of a circle.

**Theorem 1:** The interior angle between two secants drawn outside a circle is half the difference of the measure of the arcs intercepted by the secants. (Secant angle theorem)In the above diagram, PAB and PCD are two intersecting secants from point P. As per the stated theorem,

measure of angle P = (

$\frac{1}{2}$) (measure of arc BD - measure of arc AC)

For proving the theorem, let us join BC and AD. Let the intersect at E.

Considering triangles PBC and PDA,

angle PBC = angle PDA (angle subtended by the same arc AC on the circumference)

angle P is common.

Hence angle PCB = angle PAD (angle sum property)

The angle BAD = angle subtended by arc BD = (

$\frac{1}{2}$) $\times$ measure of arc BD

Hence angle PAD = 180

^{o} - (

$\frac{1}{2}$) $\times$ m of arc BD

**So, angle PCB in triangle PCB also = 180**^{o }- (**$\frac{1}{2}$) ****$\times$ measure of arc BD**Now in triangle PBC, angle PBC = angle subtended by arc AC = (

$\frac{1}{2}$) $\times$ measure of arc AC

Since, measure of angle P + measure of angle PBC + measure of angle PCB = 180

^{o},

measure of angle P + (

$\frac{1}{2}$) $\times$ measure of arc AC + 180

^{o} - (

$\frac{1}{2}$) $\times$ measure of arc BD = 180

^{o}So, measure of angle P = ($\frac{1}{2}$) (measure of arc BD - measure of arc AC)**Theorem 2:** When two secants of a circle intersect outside the circle, the product of the complete segment of the secant and the exterior segment of the secant is same for both the secants. (Intersecting secants theorem)Let us refer to the same diagram again. PB is a complete segment of secant PB and PA is the exterior segment of the same secant. In case of the secant PD, they are PD and PC respectively. As per this theorem, PA $\times$ PB = PC $\times$ PD

In the proof for the previous theorem we had shown that,

angle PBC = angle PDA (angle subtended by the same arc AC on the circumference)

angle P is common

angle PCB = angle PAD (angle sum property)

Therefore triangles PBC and PDA are similar.

Now as per the corresponding sides ratios of similar triangles,

$\frac{PB}{PD}$ =

$\frac{PC}{PA}$Therefore, PA $\times$ PB = PC $\times$ PD**Theorem 3:** When a secant and a tangent of a circle intersect outside the circle, the
product of the complete segment of the secant and the exterior segment of the secant is equal to the square of the segment of the tangent. (Secant - tangent intersection theorem)In fact, this theorem is a corollary of theorem 2. In the diagram shown above, if point B coincides with point A, then PB = PA. In other words, the secant PB becomes as a tangent PA. Therefore,

Therefore, PA $\times$ PB = PA^{2} = PC $\times$ PDThus theorem 3 is established.