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Internal bisectors of DeltaABC meet the ...

Internal bisectors of `DeltaABC` meet the circumcircle at points D, E and F then
Length of side EF is

A

`2R"cos"(A)/(2)`

B

`2Rsin((A)/(2))`

C

`2R cos ((C)/(2))`

D

`2R cos((B)/(2))cos ((C)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
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