For geometry lovers
For geometry lovers
Please look at the image.
$D$ is the image of a point reflection of $C_1$.
Prove: $CE=CB_1$
$D$ is the image of a point reflection of $C_1$.
Prove: $CE=CB_1$
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Re: For geometry lovers
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 $CB_1=CE$
$=\angle CB_1A_1+\angle B_1A_1C$(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 $CB_1=CE$
Last edited by sourav das on Tue Jun 26, 2012 11:48 pm, edited 1 time in total.
Reason: Sorry for the Typo,Thanks to Mahi
Reason: Sorry for the Typo,Thanks to Mahi
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: For geometry lovers
Shouldn't that be $C$ is on the perpendicular bisector of $B_1A_1$?sourav das wrote:and also $C$ is on the perpendicular bisector of $B_1C_1$.
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Re: For geometry lovers
Why is $2(\frac{\pi}{2}+\angle B_1DA_1)=2(\angle CB_1A_1)$?

 Posts: 461
 Joined: Wed Dec 15, 2010 10:05 am
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Re: For geometry lovers
As we're calculating directed angles $mod$ $\pi$
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )

 Posts: 4
 Joined: Wed Jun 27, 2012 1:32 am
Re: For geometry lovers
Could you explain it further, please?