suppose ABC be a triangle in which AD is the internal bisector of
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Step-by-step explanation:
Correct option is
A
AE is H.M. of b and c
B
AD=
b+c
2bc
cos
2
A
C
EF=
b+c
4bc
sin
2
A
D
the triangle AEF is isosceles.
Given, internal angle bisector of A meets BC at D,
we know length of AD is given by, AD =
b+c
2bc
cos
2
A
Given, △ADE is right angled triangle with right angle at D.
∠D=90
∘
,∠AED=90−
2
A
Using sine rule in △ADE, we get
sin∠AED
AD
=
sin∠ADE
AE
=
sin∠
2
A
DE
cos
2
A
AD
=AE=
sin
2
A
DE
Substitute the value of AD in above equation, we get
AE=
cos
2
A
b+c
2bc
cos
2
A
=
b+c
2bc
so, AE is harmonic mean of b and c.
Consider △ADE and △ADF , we know that
∠ADE=∠ADF=90
∘
∠DAE=∠DAF=∠
2
A
∴∠AED=∠AFD=90−∠
2
A
Hence △AEF is isosceles triangle.
EF=2DE=2AEsin
2
A
EF=2×
b+c
2bc
sin
2
A
=
b+c
4bc
sin
2
A
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