Question

Let P be a point on circumcircle of triangle ABC and perpendiculars PL PM and PN are drawn on the lines BC, CA and AB respectively prove that the point L, M, N are collinear

Answer 1

Answer:

angles in the same segment)Again ∠PMC =∠PLC = 90°Therefore, quadrilateral PMLC is cyclic....∠PML +∠PCL=180° (ii)

Now, ∠PAB +∠PCB = 180°and ∠PAN +∠PAB = 180°...∠PAN =∠PCB =∠PCL (iii)

From Equations (i) (ii) and (iii) we get,∠PML +∠PCL=∠PML +∠PAN=∠PML +∠PMN = 180°Hence, N, M, L are collinear. (NML line is called Simpson's line.