Prove that three common chords are concurrent Three circles intersect each other as shown.

Prove that the three common chords are concurrent.
Now the book does this by proving that the chord out of E and through M is the same for circle (2) and circle (3) which is very easy to show by using the intersecting chord theorem. However, why does this imply that all three chords are concurrent at M?
 A: You can prove this fact with the Power of a point.

###Theorem: Power of a point
Let $\omega$ be a circle centered at $O$ with radius $r$ and let $P$ be an arbitrary point. We denote the power of $P$ regarding $\omega$ with $\text{Pow}_\omega(P)$.
Consider now three cases


*

*$P\in\omega\implies \text{Pow}_\omega(P)=0$

*$P$ lies insider $\omega$

Then $$\begin{align*}\text{Pow}_\omega(P)&=\color{blue}{\vert BP\vert \cdot \vert PA\vert}\\&=\color{#A0F}{\vert DP\vert \cdot \vert PC\vert}\\ &=\color{fuchsia}{\vert EP\vert\cdot \vert PF\vert=\vert(r-\vert PO\vert)\cdot (r+\vert PO\vert)\vert}\\&=\color{fuchsia}{\vert r^2-\vert PO\vert ^2\vert}\end{align*}$$

*

*$P$ lies outside $\omega$

Then $$\begin{align*}\text{Pow}_\omega(P)&=\color{#A0F}{\vert PD\vert \cdot \vert PC\vert} \\&=\color{blue}{\vert PT\vert^2}\\ &=\color{fuchsia}{\vert PE\vert\cdot \vert PF\vert=\vert(\vert PO\vert- r)\cdot (\vert PO\vert +r)\vert}\\&=\color{fuchsia}{\vert\vert PO\vert^2-r^2\vert} \end{align*}$$

In general $\:$ $\fbox{$\text{Pow}_\omega(P)=\vert\vert PO\vert^2-r^2\vert$}$


Observation: 1. The converse of a point is also true.
Observation: 2. Some authors prefer to distinguish among positive and negative powers of a point.

Now we introduce a new concept.
Given two circles $\omega_1$ and $\omega_2$, the radical axis is the set of points $P$, such that

$$\text{Pow}_{\omega_1}(P)=\text{Pow}_{\omega_2}(P)$$

Now prove the following claim

If $\omega_1$ and $\omega_2$ intersect at two points, then the radical axis is the line connecting these points.


Back to your example, observe that $DC$ is the radical axis of the circles $(1)$ and $(3)$.  Hence $$\text{Pow}_{(1)}M=\text{Pow}_{(3)}M$$
Similarly, $AB$ is the radical axis of circles $(1)$ and $(2)$. We, therefore, infer that $$\text{Pow}_{(1)}M=\text{Pow}_{(2)}M$$ Considering these last equations we arrive at the desired conclusion that $$\text{Pow}_{(2)}M=\text{Pow}_{(3)}M$$ Thus, $M$ lies also on the radicual axis of the circles $(2)$ and $(3)$, i.e. the chords concur at $M$.
Observation: 3. A similar argument is used in order to prove that the perpendicular bisectors of a triangle concur at the circumcentre.
