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Friday 29 March 2013
Problem on Triangle(2)
ABC is an equilateral triangle let point P be inside triangle let D , E , F be feet of perpendiculars from P upon BC , CA, AB respectively Show that PD+PE+PF is independent of choice of P
if Area(XYZ) denotes the area of the triangle with vertices X,Y and Z, then Area(ABC) = Area(PAB) + Area(PAC) + Area(PBC) = 1/2 * s * (PD + PE + PF), thus PD+PE+PF is constant w.r.t P
if Area(XYZ) denotes the area of the triangle with vertices X,Y and Z,
ReplyDeletethen Area(ABC) = Area(PAB) + Area(PAC) + Area(PBC) = 1/2 * s * (PD + PE + PF), thus PD+PE+PF is constant w.r.t P