Question

In is the midpoint of BC. E is the foot of the perpendicular from A to BC (lie on BC), and F is the foot of the perpendicular from D to AC. Given that , and the area of triangle ABC is 84. Then the value of EF is


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Answer 1

Step-by-step explanation:

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