Question based on triangle inscribed in unit circle $ \bigtriangleup ABC $is inscribed in a unit circle.If angle bisectors of internal angles at A,B and C meet the circle at D,E and F respectively then value of $\frac{AD \cos\frac{A}{2}+BE \cos\frac{B}{2}+CF \cos\frac{C}{2}}{sin A+sinB+ sinC}$is ?
My attempt:AD,BE,CF are angle bisectors,so they meet at incenter of triangle but i dont know whether this incenter will be the center of unit circle or not.If this incenter is not center of unit circle,then i have no clue to proceed.
 A: The internal angle bisector of $\widehat{BAC}$ meet the circumcircle of $ABC$ in the midpoint of the $BC$ arc. So, if $O$ is the circumcenter of $ABC$, we have $\widehat{AOD}=\widehat{AOB}+\frac{1}{2}\widehat{BOC}=2\widehat{C}+\widehat{A}$ and:
$$ AD\cos\frac{\widehat{A}}{2} = 2R \sin\left(\frac{1}{2}\widehat{AOD}\right)\cos\frac{\widehat{A}}{2}=\sin(\widehat{C}+\widehat{A})+\sin(\widehat{C})$$
from which $AD\cos\frac{\widehat{A}}{2}=\sin(\widehat{B})+\sin(\widehat{C})$ and the wanted ratio is just $\displaystyle\color{red}{2}$.

A: \begin{align}
x&=\frac{|AD| \cos\tfrac{\alpha}{2}
+|BE|\cos\tfrac{\beta}{2}
+|CF|\cos\tfrac{\gamma}{2}}{\sin\alpha+\sin\beta+\sin\gamma}
=?\quad (1)
\end{align}

Consider $\triangle ADC$.
We know
that it shares 
the circumradius $R=1$
and the side $|AC|$
with $\triangle ABC$,
and we also know all the angles:
$\angle CAD=\tfrac\alpha2$, 
$\angle ADC=\beta$ (why?)
and $\angle ACD=\pi-\tfrac\alpha2-\beta$.
Hence, by the law of sines,
\begin{align}
|AD|&=
2R \sin(\pi-\tfrac\alpha2-\beta)
=
2R\sin(\tfrac\alpha2+\beta).
\end{align}
Similarly, from $\triangle BEA$ and $\triangle CFB$,
\begin{align}
|BE|&=
2R\sin(\tfrac\beta2+\gamma),
\\
|CF|&=
2R\sin(\tfrac\gamma2+\alpha).
\end{align}
Numerator of (1) is then
\begin{align}
&|AD|\cos\tfrac\alpha2
+|BE|\cos\tfrac\beta2
+|CF|\cos\tfrac\gamma2
\\
&=2R\left(
\sin(\tfrac\alpha2+\beta)\cos\tfrac\alpha2
+
\sin(\tfrac\beta2+\gamma)\cos\tfrac\beta2
+
\sin(\tfrac\gamma2+\alpha)\cos\tfrac\gamma2
\right)
\\
&=
2R\left(
\tfrac12\sin(\alpha+\beta)+\tfrac12\sin(\beta)
+
\tfrac12\sin(\beta+\gamma)+\tfrac12\sin(\gamma)
+
\tfrac12\sin(\gamma+\alpha)+\tfrac12\sin(\alpha)
\right)
\\
&=
2R\left(
\tfrac12\sin\gamma+\tfrac12\sin\beta
+
\tfrac12\sin\alpha+\tfrac12\sin\gamma
+
\tfrac12\sin\beta+\tfrac12\sin\alpha
\right)
\\
&=
2R(\sin\alpha+\sin\beta+\sin\gamma).
\end{align}
Hence, the answer is $x=2R=2$.
