Hi!
Can you please help me with this question....
Let P be a point on circumcircle of triangle ABC and perpendiculars PL, PM and PN are drawn on the lines through line segments BC, CA and AB respectively. Prove that the points L, M, N are collinear.

Monalisa

Assume, without loss of generality, that $P$ lies on the arc $AC$ (opposite to $B$). Now,

$\angle APC={180}^{\circ }-\angle B$ ( $ABCP$ cyclic quad)

But also $\angle NPL={180}^{\circ }-\angle B$ (the other angles in $NPLB$ are right angles)

so $\angle NPA=\angle CPL$.

Since $NAMP$ and $PMLC$ are cyclic quadrilaterals, we obtain

$\angle LMC=\angle LPC=\angle NPA=\angle NMA$

so $L$, $M$, $N$ collinear.

The line LMN is known as the Simson's line or pedal line of P.

Kerwin

Hi Kerwin!

Thanks for your help...I'd tried to solve it in another way, a bit long one, I would say.

I had done that since A,M,P,N are concylic,
so, < PMN = < PAN......(1)

But, PMLC is also a cyclic quad.,
so, < PML = < PCL......(2)

Also, < PAB + < PCB = 180 degrees
and < PAN + < PAB = 180 degrees
So, < PAN = < PCB = < PCL........(3)

From (1), (2) & (3), we get:

180 degrees = < PML + < PCL
= < PML + < PAN
= < PML + < PMN

Hence, NML is a line segment.

Monalisa.