Secondary Special Camp 2011: Geometry P 3
Secondary Special Camp 2011: Geometry P 3
Problem 3: Let $ABC$ be a triangle. Extend the side $BC$ past $C$, and let $D$ be the point on the extension such that $CD = AC$. Let $P$ be the second intersection of the circumcircle of $ACD$ with the circle with diameter $BC$. Let $BP$ and $AC$ meet at $E$, and let $CP$ and $AB$ meet at $F$. Prove that $D,E,F$ are collinear.
"Inspiration is needed in geometry, just as much as in poetry."  Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: Secondary Special Camp 2011: Geometry P 3
Let Extended $AP$ meets $BC$ at $D'$.
As, $AC=CD,\ \angle ADC=\angle CAD$
$\angle CPD'=180\circ \angle APC=\angle ADC$
$\angle CPD=\angle CAD$
$\Rightarrow \angle CPD=\angle CPD'$
$\therefore PC$ is the internal bisector of $\angle D'PD$
As, $\angle BPC=90\circ,\ PB$ is the external bisector of $\angle D'PD$
So, $\frac{CD'}{CD}=\frac{BD'}{BD}$
$\Rightarrow \frac{BD'}{CD'}=\frac{BD}{CD}$
As $AD',BE, CF$ concurrent, $\frac{BD'.CE.AF}{CD'.AE.FB}=1$
$\Rightarrow \frac{BD.CE.AF}{CD.AE.FB}=1 [\frac{BD'}{CD'}=\frac{BD}{CD}]$
$\therefore D,E,F$ collinear.
As, $AC=CD,\ \angle ADC=\angle CAD$
$\angle CPD'=180\circ \angle APC=\angle ADC$
$\angle CPD=\angle CAD$
$\Rightarrow \angle CPD=\angle CPD'$
$\therefore PC$ is the internal bisector of $\angle D'PD$
As, $\angle BPC=90\circ,\ PB$ is the external bisector of $\angle D'PD$
So, $\frac{CD'}{CD}=\frac{BD'}{BD}$
$\Rightarrow \frac{BD'}{CD'}=\frac{BD}{CD}$
As $AD',BE, CF$ concurrent, $\frac{BD'.CE.AF}{CD'.AE.FB}=1$
$\Rightarrow \frac{BD.CE.AF}{CD.AE.FB}=1 [\frac{BD'}{CD'}=\frac{BD}{CD}]$
$\therefore D,E,F$ collinear.
 Attachments

 Problem.JPG (17.38 KiB) Viewed 1600 times
Every logical solution to a problem has its own beauty.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)
Re: Secondary Special Camp 2011: Geometry P 3
Zzzz via,you wrote:$\frac{BD}{CD}\times \frac{CE}{AE}\times \frac{AF}{FB}=1$.So,D,E,F collinear.Why!
"Questions we can't answer are far better than answers we can't question"
Re: Secondary Special Camp 2011: Geometry P 3
Here he used Menelaus' theorem , a theorem to prove collinearity . If three points lie on the three sides of a triangle (or their extensions) and are collinear , then they have this fraction relation .tanmoy wrote:Zzzz via,you wrote:$\frac{BD}{CD}\times \frac{CE}{AE}\times \frac{AF}{FB}=1$.So,D,E,F collinear.Why!
Try not to become a man of success but rather to become a man of value.Albert Einstein