# Doubt: Prove that the circumcircles of $\Delta ABC$ and $\Delta ADE$ are tangent with $\sqrt {BC}$

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 ?

• You may use coordinate geometry too... It is lengthy though, but completely mechanical... You need not think much...just do whatever is asked – Soumyadwip Chanda Sep 25 '20 at 6:30
• In standard English, the word "doubt" is far from synonymous with "question". – Michael Hardy Sep 25 '20 at 6:31
• @MichaelHardy Indeed, this is a key feature of Indian English. – Toby Mak Sep 25 '20 at 6:42
• Great! Thank you very much! :) – cosmo5 Sep 25 '20 at 6:44
• 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. – 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.