## Construct a line through A

- Enthurelxyz
**Posts:**17**Joined:**Sat Dec 05, 2020 10:45 pm**Location:**Bangladesh-
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### Construct a line through A

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.

and miles to go before we sleep

and miles to go before we sleep

and miles to go before we sleep

- Anindya Biswas
**Posts:**200**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
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### Solution :

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.

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.

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.

**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.

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

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**John von Neumann**