Page **1** of **1**

### Dhaka Junior 2011/9

Posted: **Wed Feb 02, 2011 8:27 am**

by **Moon**

9. $ABC$ is a triangle with $AB = 4$, $BC = 5$ and $AC = 3$. $O$ is the midpoint of $BC$. A line from $O$ parallel to $AC$ meets $AB$ at $D$. $E$ is on $DO$ so that $DO = OE$ and $D$ and $E$ are on opposite sides of $BC$. Find $BE^2$.

### Re: Dhaka Junior 2011/9

Posted: **Wed Feb 02, 2011 11:31 am**

by **Mehfuj Zahir**

Answer is 13

### Re: Dhaka Junior 2011/9

Posted: **Wed Feb 02, 2011 1:55 pm**

by **ahmedulkavi**

How can we get the answer [13]?

### Re: Dhaka Junior 2011/9

Posted: **Wed Feb 02, 2011 2:47 pm**

by **Mehfuj Zahir**

BE^2=AD^2+DE^2

### Re: Dhaka Junior 2011/9

Posted: **Wed Jul 11, 2012 8:56 am**

by **jkisor**

May be BE=DC.

(BOE and DOC is similar)

DC^2=OC^2-OD^2=2.5^2-1.5^2=4

### Re: Dhaka Junior 2011/9

Posted: **Wed Aug 29, 2012 11:48 pm**

by **nafistiham**

$BD=2$

$DE=DO+OE=DO+DO=AC=3$

$BE^2=BD^2+DE^2=13$

### Re: Dhaka Junior 2011/9

Posted: **Thu Jan 10, 2013 5:10 pm**

by **Shafin**

nafistiham wrote:$BD=2$

$DE=DO+OE=DO+DO=AC=3$

$BE^2=BD^2+DE^2=13$

Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of e

### Re: Dhaka Junior 2011/9

Posted: **Fri Jan 25, 2013 12:56 pm**

by **nafistiham**

Shafin wrote:nafistiham wrote:$BD=2$

$DE=DO+OE=DO+DO=AC=3$

$BE^2=BD^2+DE^2=13$

Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of e

Sorry for being so late.

here is what you wanted.

*there is something called 'extension' :p* - junior 2011-9.png (18.07 KiB) Viewed 2077 times