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Question

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 ∠IFD has measure:

A
15o
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B
30o
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C
45o
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D
Any value 90o
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Solution

The correct option is B 30oLet ∠IBF=x,∠BID=y,∠BIF=z∴∠IBF= IBD= IDF= IFD=x using angles in the same arc have the same measure and BE is the angle bisector of ∠ABCSimilarly, ∠BID=∠BFD=y∠BIF=∠BDF=zSince points B,D,I,F are con-cyclic, ∠FBD+∠FID=180∴2x+y+z=180 ...(1)In ΔFCB,∠FCB=180−3x−yIn ΔBAD,∠BAD=180−3x−zSumming the angles A,B,C of ΔABC to 180, we have2x+2(180−3x−y)+2(180−3x−z)=180∴x+180−3x−y+180−3x−z=90∴270=5x+y+z ...(2)Subtracting (1) from (2), we have90=3x or x=30o

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