BdMO National 2013: Secondary 4
Posted: 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, $DE=BF$.
$AE\parallel CF \Leftrightarrow CF\perp BD$
$\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,$FD=EB=\frac{AB*CD}{AC}$
$FD+EF=EB+EF\Leftrightarrow DE=BF$
$\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,$FD=EB=\frac{AB*CD}{AC}$
$FD+EF=EB+EF\Leftrightarrow DE=BF$
