Question

Let P be any point on circumference of triangle ABC and perpendicualr PL, PM, PN are drawn on the segments BC, CA and AB respectively. Show points L,M, N are collinear.

Answer 1

To prove : N, M and L are collinear.
Construction: Join PA and PC.
Proof: $\angle PMA+\angle ANP={90}^{\circ }+{90}^{\circ }={180}^{\circ }$
Therefore, points A, M, P, N are conyclic.
$\therefore \angle PMN=\angle PAN$ ....(i)
(angles in the same segment)
Again $\angle PMC=\angle PLC={90}^{\circ }$
Therefore quadrilateral PMLC is cyclic.
$\therefore \angle PML+\angle PCL={180}^{\circ }$ .....(ii)
Now, $\angle PAB+\angle PCB={180}^{\circ }$
and $\angle PAN+\angle PAB={180}^{\circ }$
$\therefore \angle PAN=\angle PCB=\angle PCL$ ....(iii)
From Equations (i) (ii) and (iii) we get,
$\angle PML+\angle PCL=\angle PML+\angle PAN$
$=\angle PML+\angle PMN={180}^{\circ }$
Hence, N, M, L are collinear. (NML line is called simpson's line).