In triangle ABC, points D,E,F are on sides BC,CA,AB respectively such that AD,BE,CF are angle bisectors of triangle ABC. The lines AD,BE,CF are concurrent at I, the incenter of triangle ABC. If ∠BAC=92∘, what is the measure (in degrees) of ∠EIF?

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To find the measure of ∠EIF, we need to determine the relationship between this angle and the given information about the angle bisectors of triangle ABC.

Since AD, BE, and CF are angle bisectors, they divide the angles of triangle ABC into equal parts. This means that ∠BAC is divided into two equal angles by the angle bisector AD.

Since ∠BAC = 92∘, we can label these two equal angles as ∠BAD and ∠DAC, each measuring half of ∠BAC, which is 46∘.

Now, let's focus on triangle AEF. Since AD is the angle bisector of ∠BAF, it divides ∠BAF into two equal angles, denoted as ∠BAD and ∠DAF.

Given that triangle ABC is an isosceles triangle with ∠BAC = 92∘, ∠BAD = ∠DAC = 46∘.

By the angle angle theorem, the two triangles ABD and ACD are similar to triangle ABC. So, we can write the proportion:

AB/BD = AC/CD

Since AB = AC (isosceles triangle), we have:

AB/BD = AC/CD = 1.

Since we know that ∠BAD = ∠DAC = 46∘, triangle BDA and CDA are both isosceles triangles.

Therefore, ∠DBA = ∠DAB and ∠DCA = ∠DAC.

Now let's look at triangle DEF. We have:

∠EIF = ∠DAF - ∠DAE.

Since ∠DAF = ∠BAD = 46∘ and ∠DAE = ∠DAC = 46∘, we can substitute these values into the equation:

∠EIF = 46∘ - 46∘ = 0∘.

Therefore, the measure of ∠EIF is 0 degrees.