# Power of a Point Theorem

Given a point $P$ and a circle, pass two lines through $P$ that intersect the circle in points $A$ and $D$ and, respectively, $B$ and $C.$ Then $AP\cdot DP = BP\cdot CP.$

The point $P$ may lie either inside or outside the circle. The line through $A$ and $D$ (or that through $B$ and $C$ or both) may be tangent to the circle, in which case $A$ and $D$ coalesce into a single point. In all the cases, the theorem holds and is known as the **Power of a Point Theorem**.

When the point $P$ is inside the circle, the theorem is also known as the **Theorem of Intersecting Chords** (or the **Intersecting Chords Theorem**) and has a beautiful interpretation. When the point $P$ is outside the circle, the theorem becomes the **Theorem of Intersecting Secants** (or the **Intersecting Secants Theorem**.)

The proof is exactly the same in all three cases mentioned above. Since triangles ABP and CDP are similar, the following equality holds:

$\displaystyle\frac{AP}{CP} = \frac{BP}{DP},$

which is equivalent to the statement of the theorem: $AP\cdot DP = BP\cdot CP.$

The common value of the products then depends only on $P$ and the circle and is known as the *Power of Point $P$ with respect to the (given) circle*. Note that, when $P$ lies outside the circle, its power equals the length of the square of the tangent from P to the circle. For example, if $B=C$ so that $BP$ is tangent to the circle $AP\cdot DP=BP^{2}.$

Sometimes it is useful to employ signed segments. The convenience is that it is possible to tell points inside the circle from the points outside the circle. The power of a point inside the circle is negative, whereas that of a point outside the circle is positive. This is exactly what one obtains from the algebraic definition of the power of a point.

The theorem is reversible: Assume points $A,$ $B,$ $C,$ and $D$ are not collinear. Let $P$ be the intersection of $AD$ and $BC$ such that $AP\cdot DP = BP\cdot CP.$ Then the four points $A,$ $B,$ $C,$ and $D$ are concyclic. To see that draw a circle through, say, $A,$ $B,$ and $C.$ Assume it intersects $AP$ at $D'.$ Then, as was shown above, $AP\cdot D'P = BP\cdot CP,$ from which $D = D'.$ (If, say, $B$ and $C$ coincide, draw the circle through $A$ tangent to $PB$ at $B.)$

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