BdMO National 2013: Secondary 4

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

BdMO National 2013: Secondary 4

Unread post by BdMO » Fri Jan 10, 2014 1:36 am

$ABCD$ is a quadrilateral where $\angle B=\angle D=90^{\circ}$. $E$ and $F$ are two points on $BD$ such that $AE$ is perpendicular to $BD$ and $CF||AE$. Prove that, $DE=BF$.

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: BdMO National 2013: Secondary 4

Unread post by Tahmid » Sun Jan 12, 2014 6:19 pm

$AE\parallel CF \Leftrightarrow CF\perp BD$
$\angle B+\angle D=90+90=180\Leftrightarrow ABCD $is concyclic

now, $\Delta ABC\sim \Delta DFC$
$\Leftrightarrow \frac{FD}{CD}=\frac{AB}{AC}\Leftrightarrow FD=\frac{CD*AB}{AC}$

samely, $\Delta ADC\sim \Delta AEB$
$\Leftrightarrow \frac{EB}{AB}=\frac{CD}{AC}\Leftrightarrow EB=\frac{AB*CD}{AC}$

so,$FD=EB=\frac{AB*CD}{AC}$
$FD+EF=EB+EF\Leftrightarrow DE=BF$
......proved... :)

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