Ratio of the sides

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Fahim Shahriar
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Ratio of the sides

Unread post by Fahim Shahriar » Fri Jan 18, 2013 1:03 am

$ABCD$ is a rectangle. $O$ is the midpoint of $AC$. On $O$ point a perpendicular is drawn which intersect $AB$ in $M$ point.
$\frac {AM}{BM} = 2$

Find $\frac {AB}{BC}$.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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SANZEED
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Location:Mymensingh, Bangladesh

Re: Ratio of the sides

Unread post by SANZEED » Fri Jan 18, 2013 10:08 am

Let us assume that the perpendicular bisector of $AC$ intersect $CD$ at $N$. Let $O$ be the midpoint of $AC$.
Also assume that $MB=1$. So $AM=2$. Now by the ASA version of congruence, $\triangle AOM$ and $\triangle COM$ are congruent. Thus $CM=2$. But $\triangle CBM$ is a right triangle so by Pythagoras $BC= \sqrt {3}$ So the desired ratio is $ \sqrt {3}:1$.
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Fahim Shahriar
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Re: Ratio of the sides

Unread post by Fahim Shahriar » Fri Jan 18, 2013 12:45 pm

I did:

Let $AB=a$ & $BC=b$
$OA = \frac {\sqrt {a^2+b^2}}{2}$

In $\triangle AOM, Cos A = \frac {OA}{\frac{2a}{3}}$

In $\triangle ABC, Cos A = \frac {a}{2OA}$

Solving them we will get $\frac {a}{b} = \sqrt 3$
Last edited by Phlembac Adib Hasan on Fri Jan 18, 2013 12:47 pm, edited 1 time in total.
Reason: Latexed properly
Name: Fahim Shahriar Shakkhor
Notre Dame College

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