## Dhaka Junior 2011/9

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### Dhaka Junior 2011/9

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 $BE^2$.

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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**Posts:**14**Joined:**Tue Feb 01, 2011 11:20 am

### Re: Dhaka Junior 2011/9

How can we get the answer [13]?

### Re: Dhaka Junior 2011/9

May be BE=DC.

(BOE and DOC is similar)

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

(BOE and DOC is similar)

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

- nafistiham
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
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### Re: Dhaka Junior 2011/9

$BD=2$

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

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

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

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

\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

### Re: Dhaka Junior 2011/9

Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of enafistiham wrote:$BD=2$

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

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

- nafistiham
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
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### Re: Dhaka Junior 2011/9

Sorry for being so late.Shafin wrote:Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of enafistiham wrote:$BD=2$

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

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

here is what you wanted.

\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.