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 concurrent at I and B, D, I, F are concyclic, then ZIFD has measure:
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Answer:
Let ∠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 ∠ABC
Similarly, ∠BID=∠BFD=y
∠BIF=∠BDF=z
Since points B,D,I,F are con-cyclic, ∠FBD+∠FID=180
∴2x+y+z=180 ...(1)
In ΔFCB,∠FCB=180−3x−y
In ΔBAD,∠BAD=180−3x−z
Summing the angles A,B,C of ΔABC to 180, we have
2x+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 have
90=3x or x=30
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