$△$ ABD is inscribed in a circle.Poit P lies on the circumscribed circle of a traingle be such that through point P. PN,PM and PL are perpendicular on sides of triangle (possibly by increasing sides).Prove that points N,M,L are collinear.

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Solution

Construction :- Join NM , ML and AP

$∠ PNA = 90 ° (given) ∠ PMA = 90 ° (given) Adding both we get, ∠ PNA + ∠ PMA = 180 ° Since, the opposite angles of a quadrilateral are supplementary . Hence, PNAM is a cyclic quadrilateral . ∠ PAN = ∠ PMN . . . . (1) (angles in the same segment) PABC is acyclic quadrilateral . ∠ PAN = ∠ PCB (Exterior angle of a cyclic quadrilateral is equal to its interior opposite angle) or, ∠ PAN = ∠ PCL . . . . . . . . . . . . . . . (2) from equation (1) and (2) ∠ PMN = ∠ PCL . . . . . . . . (3) ∠ PMC = ∠ PLC = 90 ° Since, PC subtend equal angles at points M and L . Hence, PMLC is a cyclic quadrilateral . So, ∠ PCL + ∠ PML = 180 ° \Rightarrow ∠ PMN + ∠ PML = 180 ° (using eqn 3) Also, ∠ PMN and ∠ PML are adjacent angles . Hence, points N, M and L are collinear .$

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