Given $2$ circles with centres $O$ and $O^{'}$ in the Euclidean plane, one draws a couple of transverse common tangents to these circles, $T_{1}$ and $T_{2}$, and mark their intersection point as $M$. Prove that $M$ lies on $OO^{'}$.
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This is essentially homothety, consider the following proof:

Suppose the tangents intersect $OO'$ at $M_1,M_2$. Prove that $\frac {OM_1}{O'M_1}=\frac {OM_2}{O'M_2}$ and that this is sufficient to conclude that $M_1=M_2=M$.

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TopNewest@Xuming Liang @Nihar Mahajan @Daniel Liu

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This is clear by symmetry.

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This is essentially homothety, consider the following proof:

Suppose the tangents intersect $OO'$ at $M_1,M_2$. Prove that $\frac {OM_1}{O'M_1}=\frac {OM_2}{O'M_2}$ and that this is sufficient to conclude that $M_1=M_2=M$.

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Thank You !

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You could also bash the problem and prove that M lies on the equation of OO'.

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