Let P be a point on the circumcircle of DABC and perpendiculars PL, PM and PN are drawn on BC, CA and AB respectively. Show that L, M and N are collinear (lie on the same straight line).
To Prove: N, M and L are collinear.
Construction: Join PA and PC.
Proof: ∠PMA + ∠ANP = 90° + 90° = 180°
Therefore, points A, M, P, N are concyclic.
... ∠PMN = ∠PAN ................... (i)
(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).
Construction: Join PA and PC.
Proof: ∠PMA + ∠ANP = 90° + 90° = 180°
Therefore, points A, M, P, N are concyclic.
... ∠PMN = ∠PAN ................... (i)
(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).
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