Question

In a triangle ABC ,the angle bisectors BE and CF are equal ,then AB=

Answer 1

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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Answer 2

Answer:

sorry I didn't know

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