In triangle ABC,let AD, BE and CF be the inte | vedclass

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(B) Given that $AD, BE, CF$ are the internal angle bisectors of $	riangle ABC$ and they concur at the incenter $I$.Since $B, D, I, F$ are concyclic,t

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$30^{\circ}$

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(B) Given that $AD, BE, CF$ are the internal angle bisectors of $	riangle ABC$ and they concur at the incenter $I$.Since $B, D, I, F$ are concyclic,the angles subtended by the same arc $ID$ are equal.Therefore,$\angle IFD = \angle IBD$.Since $BE$ is the angle bisector of $\angle B$,we have $\angle IBD = \frac{\angle B}{2}$.In the cyclic quadrilateral $BDIF$,the sum of opposite angles is $180^{\circ}$.Thus,$\angle FBD + \angle FID = 180^{\circ}$.We know $\angle FBD = \angle B$.In $	riangle ABD$,$\angle FID$ is an exterior angle to $	riangle BDI$,so $\angle FID = \angle IBD + \angle IDB = \frac{\angle B}{2} + \angle ADB$.In $	riangle ABD$,$\angle ADB = 180^{\circ} - (\angle BAD + \angle ABD) = 180^{\circ} - (\frac{A}{2} + B)$.So,$\angle FID = \frac{B}{2} + 180^{\circ} - \frac{A}{2} - B = 180^{\circ} - \frac{A+B}{2} = 180^{\circ} - \frac{180^{\circ}-C}{2} = 90^{\circ} + \frac{C}{2}$.Substituting into the cyclic condition: $B + 90^{\circ} + \frac{C}{2} = 180^{\circ} \Rightarrow B + \frac{C}{2} = 90^{\circ}$.Since $I$ is the incenter,$\angle BID = 180^{\circ} - (\frac{B}{2} + \angle IDB)$.Alternatively,using the property of the cyclic quadrilateral $BDIF$,$\angle IFD = \angle IBD = \frac{B}{2}$.From the condition $\angle FBD + \angle FID = 180^{\circ}$,we have $B + (180^{\circ} - \angle BDI) = 180^{\circ}$ is not correct. The correct condition is $\angle IFB + \angle IDB = 180^{\circ}$.Given $BDIF$ is cyclic,$\angle IFD = \angle IBD = \frac{B}{2}$ and $\angle IDF = \angle IBF = \frac{B}{2}$.In $	riangle BDF$,$\angle BFD = 180^{\circ} - (B + \frac{B}{2} + \frac{B}{2}) = 180^{\circ} - 2B$.Also $\angle IFD = \frac{B}{2}$. For this configuration,$B=60^{\circ}$ is required.Thus,$\angle IFD = \frac{60^{\circ}}{2} = 30^{\circ}$.

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 $\angle IFD$ has measure $......$

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