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 know the relationship between the angles in triangle ABC and triangle DEF.
First, let's find the measure of ∠B. Since AD is an angle bisector, we know that ∠BAD = ∠CAD. Therefore, ∠B = ∠BAD + ∠CAD = (∠BAD + ∠CAD) / 2 = (∠BAC) / 2 = 92° / 2 = 46°.
Now, let's find the measure of ∠E. Since BE is an angle bisector, we know that ∠EBF = ∠EAF. Therefore, ∠E = ∠EBF + ∠EAF = (∠EBF + ∠EAF) / 2 = (∠B + ∠C) / 2 = (46° + ∠C) / 2.
Finally, to find the measure of ∠EIF, we need to find the measure of ∠FIN. Since AD is an angle bisector, we know that ∠FIA = ∠EIA. Therefore, ∠EIF = 180° - ∠FIA - ∠FIN = 180° - (∠FIA + ∠EIA) = 180° - (2 * ∠FIA) = 180° - 2 * (∠B / 2) = 180° - 2 * 46° = 180° - 92° = 88°.
Therefore, the measure of ∠EIF is 88 degrees.