Half area

From a problem in Hyacinthos

Given a triangle ABC, with circumcenter O, and let P any point on side BC, M and N the perpendicular projection of P on AC and AB.

The area of AMON is half area of ABC

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Solution by Nikos Dergiades :

Recall a few relations :
In a triangle ABC, R the circumcircle radius, a,b,c the sides and A,B,C the angles : 2R = a/sin(A) = b/sin(B) = c/sin(C) = a.b.c/2S

Area(ABC) = S = 1/2 b.c.sin(A) = ...

Therefore, in a quadrilateral with diagonals x, y at angle θ : Area = 1/2 x.y.sin(θ)
(by decomposition in 4 triangles)
And after this recall :

Let AH the altitude in ABC, AH' the altitude in AMN
AP is the diameter of the circumcircle to AMPN, and is isogonal conjugate of AH' :
(AN,AP) = (MN,MP) = (AH',AM)
As (AN,AH) = (AO,AM) (AH and AO isogonal conjugate), (AO,AH') = (AP, AH) and (AO,MN) = (AP,BC) = θ
Area of AMON is 1/2 AO.MN.sin(θ)
That is, with AO = R and MN = AP.sin(A) (AP = diameter of circumcircle to AMN), Area(AMON) = 1/2 R.AP.sin(A).sin(θ)
Area of ABC is 1/2 AP.BC.sin(θ)     ("quadrilateral" ABPC, with diagonals AP and BC)
As BC = 2R.sin(A), Area(ABC) = R.AP.sin(A).sin(θ)
QED.

 

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