## BdMO National 2013: Secondary 7

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
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### BdMO National 2013: Secondary 7

$ABCD$ is s quadrilateral. $AB||CD$. $P$ is a point on $AB$ and $Q$ is a point on $CD$. A line parallel to $AB$ intersects $AD$, $BC$, $DP$, $CP$, $AQ$, $BQ$ at points $M, N, X, Y, R, S$ respectively. Prove that $MX+NY=RS$.

Siam
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### Re: BdMO National 2013: Secondary 7

Is the question correct? Will it be $MX+NY=RS$ or $MY+NX=RS$ ?

*Mahi*
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### Re: BdMO National 2013: Secondary 7

There was a printing mistake in last year's question, $CP , DP$ would be $DP, CP$. It's been edited accordingly now, thanks for pointing out
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Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Siam
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### Re: BdMO National 2013: Secondary 7

Since $MX \| AP$, $\triangle MXD \sim \triangle APD$
Similarly, $\triangle RSQ \sim \triangle AQB$ and$\triangle NYX \sim \triangle NPB$
Let the perpendicular distance between $AB$ and $CD$ be $h$ and $MN$ to $CD$ be $i$
Now, we have, $[APD]+[PBC]=.5(h \times AP)+(h \times PB)=.5 \times h\times (AP+BP)=[AQB]$
Let $\frac{h}{i} =k$
So we have, $[APD]=k^{2} [MXD]$, $[AQB]=k^{2} [RSQ]$, $[PBC]=k^{2} [NCY]$

So, $k^{2} [MXD]+k^{2} [NYC]=k^{2} [RSQ]$
Or, $[MXD]+[NYC]=[RSQ]$
Or, $i \times MX+i \times NY=i \times RS$
Or, $MX+NY=RS$