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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. :oops: :oops:
here is what you wanted.
junior 2011-9.png
there is something called 'extension' :p
junior 2011-9.png (18.07 KiB) Viewed 2077 times