Question

prove that points N, M L are collinear​

Answer 1

Answer:

Given data:

Point P lies on the circumcircle of ∆ ABC.

PL ⊥ BC

PN ⊥ AB extended

PM ⊥ AC

To prove: N, M, & L are collinear

Proof:

Step 1:

Let’s join PA, PC, MN & ML.

Since PN ⊥ AB & PM ⊥ AC

∴ ∠PMA = ∠PNA = 90°

⇒ Points A, M, N & P are concyclic

So, AMPN is a cyclic quadrilateral

Since angle PMN & angle PAN are the angles lying in the same segment of the chord PN

∴ ∠PMN = ∠PAN …… (i)

Step 2:

Since PL ⊥ BC & PM ⊥ AC

∴ ∠PMC = ∠PLC = 90°

Since the line joining the P and C subtends equal angles at points M & L lying on its same side

∴ Points P, C, L & M are concyclic

So, PCLM is a cyclic quadrilateral

∴ ∠PML +∠PCL = 180° ….. [∵ sum of opposite pairs of angles of a cyclic quadrilateral is 180°]…. (ii)

Step 3:

Since ABCP is cyclic quadrilateral and BAN is a straight line.

∴ ∠PAB + ∠PCB = 180° ….. [∵ sum of opposite pairs of angles of a cyclic quadrilateral is 180°]…. (iii)

&

∠PAB + ∠PAN = 180° …… [straight angle]….. (iv)

From (iii) & (iv), we get

∠PAB + ∠PCB = ∠PAB + ∠PAN

⇒ ∠PCB = ∠PAN

⇒ ∠PCL = ∠PAN ….. [∵ ∠PCB = ∠PCL] ….. (v)

Step 4:

From (i) & (v), we get

∠PMN = ∠PCL …… (vi)

From (ii) & (vi), we get

∠PML + ∠PMN = 180°

∴ NML is a straight line

Thus, points N, M, & L are collinear .