[asy][asy] unitsize(3cm); draw(circle((0,0),1)); draw(dir(160) -- dir(70)); draw(dir(20) -- dir(140)); draw(dir(270) -- dir(120)); draw(dir(90) -- dir(180)); draw(dir(180) -- dir(20) -- dir(120) -- dir(160) -- dir(180)); label("$A$",dir(120), N); label("$B$",dir(160), W); label("$C$",dir(180), W); label("$D$",dir(20), E); label("$M$",dir(140), W); label("$N$",dir(70), N); label("$P$",dir(90), N); label("$Q$",dir(270), S); draw(dir(160) -- dir(20)); label("$I$", dir(130)*0.5, N); [/asy][/asy] Suppose $DI,BI,CI,AI$ intersect the circle at $M,N,P,Q$ respectively. Start by doing some angle chasing: $$\angle MDC = \angle IDC = \angle BIC = 180^\circ - \angle BCP - \angle NBC. $$Then $$\overarc{$MBC$} = 180^\circ - \overarc{$BP$} - \overarc{$NQC$}.$$Hence $$\overarc{$BP$} = 180^\circ - \overarc{$MBC$} - \overarc{$NQC$} = \overarc{$MN$}.$$On the other hand, by similarity, we have $$\dfrac{AM}{DQ} = \dfrac{AI}{DI}, \dfrac{DN}{BM}=\dfrac{DI}{BI},\dfrac{BQ}{AN}=\dfrac{BI}{AI}.$$Because $AI$ bisects $\angle BAD,$ $BQ=QD.$ Multiply these equations: $$\dfrac{AM}{DQ} \dfrac{DN}{BM}\dfrac{BQ}{AN}=\dfrac{AM}{AN} \dfrac{DN}{BM}=1.$$If $AM>BM,$ then $$\overarc{$AM$}>\overarc{$BM$}.$$Because $CP$ bisects $\angle BCD$, $$\overarc{$AN$}<\overarc{$DN$}.$$Hence $\dfrac{AM}{AN} \dfrac{DN}{BM}>1.$ Similarly we can show that it is impossible that $AM<BM.$ Thus $AM=BM$ and $I$ is the incenter.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -27.876520371111255, xmax = 27.058275043871003, ymin = -16.71227689947088, ymax = 17.612993961132382; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((2.284,6.278)--(-4.782,-1.24)--(6.338,1.04)--cycle, linewidth(2) + zzttqq); draw((6.338,1.04)--(2.108144524261927,2.607797024884341)--(5.492817515943216,-2.2250378477853023)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((2.284,6.278)--(-4.782,-1.24), linewidth(0.4) + red); draw((-4.782,-1.24)--(6.338,1.04), linewidth(0.4) + red); draw((6.338,1.04)--(2.284,6.278), linewidth(0.4) + red); draw(circle((0.5966916755272806,0.7842756877792273), 5.747000626459035), linewidth(0.4)); draw((-0.5576361412228421,6.4141552150868435)--(5.492817515943216,-2.2250378477853023), linewidth(0.4)); draw((-4.782,-1.24)--(5.1414961079806005,4.301770374523359), linewidth(0.4)); draw((6.338,1.04)--(-3.5909877249415154,4.7201818654478265), linewidth(0.4)); draw((6.338,1.04)--(2.108144524261927,2.607797024884341), linewidth(0.4) + blue); draw((2.108144524261927,2.607797024884341)--(5.492817515943216,-2.2250378477853023), linewidth(0.4) + blue); draw((5.492817515943216,-2.2250378477853023)--(6.338,1.04), linewidth(0.4) + blue); draw(circle((3.5539946019787774,0.018752974104690016), 2.965405797544633), linewidth(0.4) + blue); /* dots and labels */ dot((2.284,6.278),dotstyle); label("$A$", (2.4273798447351713,6.626034878135939), NE * labelscalefactor); dot((-4.782,-1.24),dotstyle); label("$B$", (-4.645923878631826,-0.8781299857407152), NE * labelscalefactor); dot((6.338,1.04),dotstyle); label("$D$", (6.484655584534515,1.3838910019350228), NE * labelscalefactor); dot((2.108144524261927,2.607797024884341),linewidth(4pt) + dotstyle); label("$I$", (2.2478543695228113,2.891904993718848), NE * labelscalefactor); dot((-0.5576361412228421,6.4141552150868435),linewidth(4pt) + dotstyle); label("$M$", (-0.4091226636201217,6.697845068220883), NE * labelscalefactor); dot((5.492817515943216,-2.2250378477853023),linewidth(4pt) + dotstyle); label("$C$", (5.622933303515185,-1.955282837014876), NE * labelscalefactor); dot((5.1414961079806005,4.301770374523359),linewidth(4pt) + dotstyle); label("$P$", (5.299787448132936,4.579444460715034), NE * labelscalefactor); dot((-3.5909877249415154,4.7201818654478265),linewidth(4pt) + dotstyle); label("$Q$", (-3.461055742230247,5.010305601224698), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice and easy First of all we define $M$ to be the intersection of the circle $(ABD)$ and line $CI$, by this way we have that $M$ is the midpoint of the arc $BD$.Also let $Q$ be the intersection of $DI$ with the circle $(ABD)$ and let $P$ be the intersection of $BI$ with $(ABD)$. Since we have that $\angle BIC = \angle IDC$, this implies that the circumcircle of $DIC$ is tangent to $BI$. Thus we have that: $$\angle DIP = \angle DCI = \angle DCM = \angle DBM = 90-\frac{1}{2}\angle BAD$$this implies that $\angle BID = 90+\frac{1}{2}\angle BAD$, and since we have that $I$ is on the angle bisector of $\angle BAD$, this implies that $I$ is the incenter of $BAD$.

Let $X=IB\cap CD$ (maybe at infinity) Then $\angle{BID}=\angle{XID}=\angle{XCI} =\hat{A}+\frac12\hat{C}=90+\frac12\hat{A}.$ As $I$ lies on the bisector of $\hat{A}$ we get the desired result.