$ABCD$ 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 $\frac {AB}{BC}$.
Ratio of the sides
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Ratio of the sides
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$.
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$.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Ratio of the sides
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 $\frac {a}{b} = \sqrt 3$
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
Reason: Latexed properly
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College