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Let $A$ be one of the common points of two intersecting circles. Through $A$ construct a line on which the two circles cut out equal chords.

Let's name the circles $\Gamma_1$ and $\Gamma_2$. Let's assume $\Gamma_1\cap\Gamma_2=\{O,A\}$. Let's construct line $PQ$ such that $A\in PQ, PQ\cap\Gamma_1=P, PQ\cap\Gamma_2=Q$. Let $M$ be the midpoint of the segment $PQ$. Let's construct line $l$ through $O$ such that $\measuredangle (OP, l)=\measuredangle MOA$. Let $X=l\cap\Gamma_1$. The line $XA$ is the line on which $\Gamma_1$ and $\Gamma_2$ cut out equal chords.

Proof :
Let $Y=\Gamma_2\cap XA$. By Spiral Similarity, $OXAY\sim OPMQ$. Since $M$ is the midpoint of segment $PQ$, $A$ must be the midpoint of segment $XY$. Which completes our proof.

Another approach using Homothety :
Note that we could also solve it in this way,
Let's construct a circle $\Omega$ such that $\text{Radius}(\Gamma_1)=\text{Radius}(\Omega)$, $\Omega$ is externally tangent to $\Gamma_1$ at $A$. Let $Y=\Gamma_2\cap\Omega$. Then $YA$ is our wanted line.

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