Prove that three common chords are concurrent

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.