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The internal bisector of angle A of Delt...

The internal bisector of `angle A` of `Delta ABC` meets BC at D and the external bisector of `angle A` meets BC produced at E. Prove that `(BD)/(BE) = (CD)/(CE) ` .

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

`DeltaAEF` is isosceless

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Since `Delta ABC=DeltaABD+DeltaACD`
`rArr (1)/(2) bc "sin" A=(1)/(2) c AD "sin" (A)/(2) +(1)/(2) b AD "sin" (A)/(2)`

`rArr D=(2ab)/(a+c) "cos"(A)/(2)`
Again `AE=AD "sec" (A)/(2)=(2bc)/(b+c)`
` rArr AE ` HM of b and c
`EF=ED+DF=2DE=2AD "tan"(A)/(2) =(4bc)/(b+c)"sin"(A)/(2)`
Since, `ADbotEF and DE=DF and AD` is bisectors.
` rArr DeltaAEF` is isoscelers.
Hence, (a),(b),(c),(d) are correct answers
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