KVPY SA Exam 2012 Question Paper

Question : 6 of 80

Marks: +1, -0

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 1 and B, D, I, F are concyclic, then ∠IFD has measure

15°
30°
45°
any value ≤ 90°

Validate

Solution:

∠AIC=180°-({A+C}/2)

=90°+B/2

Here :-

90+B/2+B=180°

B = 60º
This will be case of equilateral ∆
⇒∠IFD = 30º

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