IIT JEE , AIEEE Forum - Mathematics , Algebra

md_2674062

IF ABC is right triangle at A .And angle bisector BD of B(D on AC) and CE bisector of C(E on AB) meet at I.
PROVE THAT AREA OF 1/2 BCDE = OF IBC

sboosy

let the usual symbols stand for what they are
that is sides of the triangle are a,b,c
now rt angled at A means a*a=b*b+c*c .....1
the two mentioned lines meet at I the symbol usually used
now area of triangle IBC
drop a perpendicular IK on BC(actually the inradius)
now required area = area of IKB+area of IKC
IK=r,KB=rcot(B/2),KC=rcot(C/2)
total area = 1/2 * r * r *(cotB/2 +cotC/2)
using r= area of abc / semi perimeter we get it equal to
b*b*c*c/(a+b+c)*(a+b+c)
cotB/2=(a+c)/b
cotC/2=(a+b)/c
thus we get area of IBC
now area of bcde=
area of ABC - area of triangle ADE
area ABC = 1/2 *b* c
area of ADE =
1/2 * AD *AE
using angle bisector theorem
a/b=BE/EA
thus EA=bc/(a+b)
similarly AD=bc/(a+c)
now substitue
for simplification use the following
(a+b+c)*(a+b+c)=2[b*b+c*c+ab+bc+ca]
bcos of 1
thus the result
....nice question

md_2674062

THANKS BRO HATS OF TO U

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