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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^{'} $. .

#Geometry #Proof #IGuessEuclidShallHelpMe #HaveNoIdea

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

@Xuming Liang @Nihar Mahajan @Daniel Liu

Venkata Karthik Bandaru - 5 years, 9 months ago

This is clear by symmetry.

Daniel Liu - 5 years, 9 months ago

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 $M_1=M_2=M$ .

Xuming Liang - 5 years, 9 months ago

Thank You !

Venkata Karthik Bandaru - 5 years, 9 months ago

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

Aditya Kumar - 5 years, 9 months ago

@Aditya Kumar – Yeah, but I don't like it. I only like Euclid-style proofs, they are more beautiful.

Venkata Karthik Bandaru - 5 years, 9 months ago

@Venkata Karthik Bandaru – Yes u r right euclid style proofs r really awsome. but they require high level intuition. bashing doesn't require much intuition.

Aditya Kumar - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$