Source: Mathcenter Contest / Oly - Thai Forum 2008 R1 p7 https://artofproblemsolving.com/community/c3196914_mathcenter_contest
Tags: geometry, area of a triangle, areas, geometric inequality
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$ABC$ is a triangle with an area of $1$ square meter. Given the point $D$ on $BC$, point $E$ on $CA$, point $F$ on $AB$, such that quadrilateral $AFDE$ is cyclic. Prove that the area of $DEF \le \frac{EF^2}{4 AD^2}$.
(holmes)
Take point $G$ on $AC$ such that $DG\parallel AB$.
Note that $\Delta DEF\backsim\Delta GDA(A.A)\Rightarrow \frac{S_{DEF}}{S_{GDA}} =\frac{EF^2}{AD^2}$.
We have: $S_{GDA} = S_{ADC} . \frac{AG}{AC} = S_{ABC} . \frac{CD}{BC} . \frac{AG}{AC} = \frac{BD}{BC} . \frac{CD}{BC}\leq \frac{1}{4}\left(\frac{BD}{BC} + \frac{CD}{BC}\right)^2 = \frac{1}{4}$
$\Rightarrow S_{DEF} \leq \frac{EF^2}{4AD^2}$.
Attachments:
Hoanglahoang wrote:
Take point $G$ on $AC$ such that $DG\parallel AB$.
Note that $\Delta DEF\backsim\Delta GDA(A.A)\Rightarrow \frac{S_{DEF}}{S_{GDA}} =\frac{EF^2}{AD^2}$.
We have: $S_{GDA} = S_{ADC} . \frac{AG}{AC} = S_{ABC} . \frac{CD}{BC} . \frac{AG}{AC} = \frac{BD}{BC} . \frac{CD}{BC}\leq \frac{1}{4}\left(\frac{BD}{BC} + \frac{CD}{BC}\right)^2 = \frac{1}{4}$
$\Rightarrow S_{DEF} \leq \frac{EF^2}{4AD^2}$.
nice solution!
