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So, I recently started inversion and I have doubt in this solution . It's from "A beautiful Journey through Olympiad geometry " by Stefan Lozanovski. This Problem uses $\sqrt{BC}$

Here , I couldn't understand this line

Furthermore, the circles $\omega _1$ and $\omega_2$ are tangent to $(ABC)$ and to the parallel lines $\ell'$ and $BC$, so they are symmetric with respect to the perpendicular bisector of $BC$.Thus, $D'$ and $E'$ are symmetric with respect to the perpendicular bisector of $BC$ as well.

I understood that it's enough to show that $D'E'\parallel BC $ , and I also observed that $\omega _1$ and $\omega_2$ are congruent . But the above lines aren't clear , can someone explain it ?

Thanks in advance !

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    $\begingroup$ You may use coordinate geometry too... It is lengthy though, but completely mechanical... You need not think much...just do whatever is asked $\endgroup$ – Soumyadwip Chanda Sep 25 '20 at 6:30
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    $\begingroup$ In standard English, the word "doubt" is far from synonymous with "question". $\endgroup$ – Michael Hardy Sep 25 '20 at 6:31
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    $\begingroup$ @MichaelHardy Indeed, this is a key feature of Indian English. $\endgroup$ – Toby Mak Sep 25 '20 at 6:42
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    $\begingroup$ Great! Thank you very much! :) $\endgroup$ – cosmo5 Sep 25 '20 at 6:44
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    $\begingroup$ I think it's saying that $\omega_1^{'}$ and $\omega_2^{'}$ are congruent circles between two parallel lines. So they are symmetric wrt any chord parallel to these two lines. To see this, just draw perpendicular bisector of BC. So $BT'_{2}=CT'_{1}$ where $T'_{i}$ is tangency point of $\omega_i^{'}$ wrt BC. $\endgroup$ – cosmo5 Sep 25 '20 at 7:06

It's basically Thales' theorem. Let $O_1,O_2,O$ be centers of $\{\omega_1', \omega_2',\odot(ABC)\}$. Thus, $O_1-D'-O$ and $O_2-E'-O$ are collinear. Trivially, $\omega_1'$ and $\omega_2'$ have same radius say $r$. Thus, $OO_1=OO_2=r+R$ where $R$ is radius of $\odot(ABC)$. Further, $OD'=OE'=R\overset{\text{Thales'}}{\implies} D'E'\|O_1O_2$ but $O_1O_2\| BC\implies D'E'\| BC$ so done.


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