Let f,g,h be the length of the perpendiculars from the circumcentre of the ΔABC on the sides a, b, and c, respecitively then the value of k for which af+bg+ch=kabcfgh, is
14
Distance of circumcenter from to side BC is = RcosA=f
Similiarly g=RcosB,h=RcosC⇒af+bg+ch=2RsinARcosA+2RsinBRcosB+2RsinCRcosC=2(tanA+tanB+tanC)
Also afbgch=8tanA tanB tanC
But in a triangle
tanA+tanB+tanC=tanA tanB tanC⇒af+bg+ch=14abcfgh