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In triangle $\mathrm{ABC}$, let $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ be the internal angle bisectors with D, E and $\mathrm{F}$ on the sides $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$ respectively. Suppose $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ concur at $\mathrm{I}$ and $\mathrm{B}, \mathrm{D}, \mathrm{I}, \mathrm{F}$ are concyclic, then $\angle \mathrm{IFD}$ has measure -
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Verified Answer
The correct answer is:
$30^{\circ}$
$\begin{array}{l}
\angle \mathrm{AIC}=180^{\circ}-\left(\frac{\mathrm{A}+\mathrm{C}}{2}\right) \\
=180^{\circ}-\left(\frac{\mathrm{A}+\mathrm{C}}{2}\right) \\
=90+\frac{\mathrm{B}}{2}
\end{array}$
Here $:-90+\frac{\mathrm{B}}{2}+\mathrm{B}=180^{\circ}$
$\mathrm{B}=60^{\circ}$
This will be case of equilateral $\Delta$ $\Rightarrow \angle \mathrm{IFD}=30^{\circ}$
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