BdMO Online Forum

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.

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.

Zzzz via,you wrote:$\frac{BD}{CD}\times \frac{CE}{AE}\times \frac{AF}{FB}=1$.So,D,E,F collinear.Why!

tanmoy wrote:Zzzz via,you wrote:$\frac{BD}{CD}\times \frac{CE}{AE}\times \frac{AF}{FB}=1$.So,D,E,F collinear.Why!

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 .