IIT JEE Advanced 2006 Paper 1

The condition is depicted in the following figure:

Now, Area of ΔABC = Area of ΔABD + Area of ΔACD
(sina×

1

2

×bc) = (

1

2

×c×AD×sin

A

2

) + (

1

2

×b×AD×sin

A

2

)
sin A (

bc

b+c

) = AD sin

A

2

AD =

2bcsin( A 2 )cos( A 2 )

(b+c)sin( A 2 )

= (

2bc

b+c

)×(cos

A

2

)
From ΔADE, we have
cos (

A

2

) =

AD

AE

⇒ AE =

2bc

b+c

where AE is HM of b and c.
From ΔADF, we have
cos (

A

2

) =

AD

AF

⇒ AR = AF ⇒ ΔAEF
which is an isosceles triangle. Therefore,
ain (

A

2

) =

DF

AF

⇒ EF = 2DF = 2AF sin

A

2

=

4bc

b+c

sin(

A

2

)