Internal bisector of <A of a triangle ABC meets side BC at D.
A line drawn through D perpendiuclar to AD intersects the side AC at E and the side AB at F.
If a,b,c represent sides of ABC then
A) AE is HM Of b and c
B) AD=2bc/(b+c). cos (A/2)
C) EF=4bc/(b+c) sin (A/2)
D) The triangle AEF is isosceles.
11
BD/CD = c/b (angular bisector of A divides side BC in the corresponding ratio of the sides)
Adding 1 to both sides, we get ,
a/CD = (b+c)/b
\Rightarrow CD = ab/(b+c)
\Rightarrow BD = ac/(b+c)
In ΔABD,
AD/(sin B) = BD/(sin (A/2))
\Rightarrow AD = [ (ac/(b+c)) / sin (A/2) ] . sin B
\Rightarrow AD = [ ac sin B / (b+c) sin (A/2) ]
\Rightarrow AD = [ bc sin A / (b+c) sin (A/2) ] [ Since area(ΔABC) = (1/2) ab sin C = (1/2) bc sin A = (1/2) ca sin B ]
Therefore AD = [ 2 bc/(b+c) ] . cos (A/2)
11
In ΔADE,
<AED = 90° - A/2
In ΔADF,
<AFD = 90° - A/2
Hence in ΔAEF,
<E = <F = 90° - A/2
Therefore ΔAEF is isosceles
11In ΔADE,
AD/AE = cos (A/2)
\Rightarrow AE = AD / cos (A/2)
\Rightarrow AE = 2bc/(b+c)
Therefore AE is HM of b and c
11
In ΔADE,
ED/AD = tan (A/2)
\Rightarrow ED = AD tan (A/2)
\Rightarrow ED = [ 2bc/(b+c) ] . sin (A/2)
In ΔAEF,
EF = 2.ED [ Since D is mid-pt. of EF as AD is altitude & ΔAEF is isosceles ]
Therefore EF = [ 4bc/(b+c) ] . sin (A/2)