## For geometry lovers

For students of class 11-12 (age 16+)
Zan
Posts: 16
Joined: Fri Jun 08, 2012 8:26 pm

### For geometry lovers

Please look at the image.
$D$ is the image of a point reflection of $C_1$.
Prove: $CE=CB_1$
Attachments 3.pdf
(13.9 KiB) Downloaded 236 times

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### 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 $CB_1=CE$
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
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

### Re: For geometry lovers

sourav das wrote:and also $C$ is on the perpendicular bisector of $B_1C_1$.
Shouldn't that be $C$ is on the perpendicular bisector of $B_1A_1$?
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Zan
Posts: 16
Joined: Fri Jun 08, 2012 8:26 pm

### Re: For geometry lovers

Nice solution!

CaptainPrice
Posts: 4
Joined: Wed Jun 27, 2012 1:32 am

### Re: For geometry lovers

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

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### 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......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

CaptainPrice
Posts: 4
Joined: Wed Jun 27, 2012 1:32 am

### Re: For geometry lovers

Could you explain it further, please?