## BDMO NATIONAL JUNIOR 2012/09

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Joined: Mon Jan 23, 2017 10:32 am

### BDMO NATIONAL JUNIOR 2012/09

Hi, guys...help me to solve this......
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Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: BDMO NATIONAL JUNIOR 2012/09

It's confusing to use $R$ for both a vertex of $PQRS$ and the cirum-radius of $\triangle ABC$.So,we will use $r$ for
the cirum-radius of $\triangle ABC$.
For the proof , we will prove two things :
$(i)\frac{AS}{PS} = \frac{c}{a}$
$(ii)\frac{PS}{SB} = \frac{b}{2r}$

$(i)$
By some angle chasing,$\angle ASR = \angle B , \angle ARS = \angle C$.
Thus,$\triangle ASR \sim \triangle ABC$.
Thus,$\frac{AS}{c} = \frac{SR}{a} \Rightarrow \frac{AS}{SR} = \frac{c}{a} \Rightarrow \frac{AS}{PS} = \frac{c}{a}$

$(ii)$
From the extended sine law , we can write : $\frac{b}{2r} = \sin B$
$\triangle BPS$ is a right-angled triangle.So, $\frac{PS}{BS} = \sin B$
Thus , $\frac{PS}{BS} = \frac{b}{2r}$.

At last,
$\frac{AS}{PS} \times \frac{PS}{SB}= \frac{c}{a} \times \frac{b}{2r}$
or, $\frac{AS}{SB} = \frac{bc}{2ar}$

$\therefore \frac{AS}{SB} = \frac{bc}{2ar}$ (proved) "(To Ptolemy I) There is no 'royal road' to geometry." - Euclid