In triangle ABC, AC is greater than AB, the internal angle bisector of angle A meets BC at D and E is the foot of perpendicular from B on AD suppose AB = 5 and BE = 4 the value of [(AC+AB)/(AC-AB)]×ED
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Answer:
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.
Step-by-step explanation:
∠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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