For geometry lovers

[Zan]

Please look at the image.
$D$ is the image of a point reflection of $C_1$.
Prove: $CE=CB_1$

[sourav das]

We're using directed angles modulo $\pi$ . $2\angle B_1EA_1=2\angle B_1ED=2(\angle EB_1D+\angle B_1DE)=2(\frac{\pi}{2}+\angle B_1DA_1)=2(\angle CB_1A_1)$ (As, $CB_1$ is tangent )
$=\angle CB_1A_1+\angle B_1A_1C$(As,$CB_1=CA_1$) $=\angle B_1CA_1$ and also $C$ is on the perpendicular bisector of$B_1A_1$ . Thus $C$ is the $Circumcenter$ of $\triangle B_1A_1E$. It implies $CB_1=CE$

[*Mahi*]

and also $C$ is on the perpendicular bisector of $B_1C_1$.

Shouldn't that be $C$ is on the perpendicular bisector of $B_1A_1$?

[Zan]

Nice solution!

[CaptainPrice]

Why is $2(\frac{\pi}{2}+\angle B_1DA_1)=2(\angle CB_1A_1)$?

[sourav das]

As we're calculating directed angles $mod$ $\pi$

[CaptainPrice]

Could you explain it further, please?