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In triangle ABC, let AD, BE and CF be th...

In triangle ABC, let AD, BE and CF be the internal angle bisectors with D, E and F on the sides BC, CA and AB respectively. Suppose AD, BE and CF concur at I and B, D, I, F are concyclic, then `angleIFD` has measure-

A

`15^(@)`

B

`30^(@)`

C

`45^(@)`

D

Any value `le90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B

`angleAIC=180^(@)-((A+C)/(2))`
`180^(@)-((A+C)/(2))`
`=90^(@)+(B)/(2)`
Here:- `=90^(@)+(B)/(2)+B=180^(@)`
`B=60^(@)`
This will be case of equilateral `Delta`
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