Let $M$ be the midpoint of $AC$ and $G$ be the centroid of $\triangle ABC$. Then $G$ lies on the median through $C$ as well as on the angle bisector of $\angle DCB = \angle ACB$, since it is also the incenter of $\triangle BCD$. Thus, we must have $AC=BC$. Moreover, since the line $\overline{BGM}$ bisects $\angle CBD$, we get $CM = \frac{CB \cdot CD}{BC+BD}$ and also $CM = AC/2 = BC/2$. Consequently, we must have $$2CD=BC+BD \Longrightarrow BC^2=BD^2+CD^2=(2CD-BD)^2 \Longrightarrow CD(3CD-4BD)=0 \Longrightarrow CD:BD=4:3.$$From this we easily get $\cos \gamma = CD/BC = \frac{4}{5}$ and hence by cosine law $$\frac{4}{5} = \frac{2BC^2-AB^2}{2BC^2} = 1-\frac{1}{2BC^2} \Longrightarrow AC=BC = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}.$$

Let $P$ be the the incenter of $\triangle BCD$ and the centroid of $\triangle ABC$. Let $K$ be the midpoint of $AC$. Note that the angle bisector of $\angle BCD$ and $C$-median of $\triangle ABC$ coincide, hence $AC=BC$. Hence $CP$ is the altitude of $\triangle ABC$. Now, let $\angle ABC=\alpha$. By angle bisector theorem, $$1-\frac{\cos{\alpha}}{\frac{0.25}{\cos{\alpha}}}=1-\frac{AD}{KC}=\frac{KC-AD}{KC}=\frac{KD}{KC}=\frac{BD}{BC}=\frac{\sin{\alpha}}{\frac{0.5}{\cos{\alpha}}},$$which is equivalent to \begin{align*} 0.5-2\cos^2{\alpha}=\sin{\alpha}\cos{\alpha}\\ 5\cos^4{\alpha}-3\cos^2{\alpha}+0.25=0\\ \cos^2{\alpha}=\frac{3\pm\sqrt{9-5}}{10}=\frac{3\pm 2}{10}. \end{align*}Thus, $\cos{\alpha}=\dfrac{\sqrt{2}}{2}$ or $\cos{\alpha}=\dfrac{1}{\sqrt{10}}$. Note that the former is impossible, because then $\angle ACB=90^\circ$ and thus $C\equiv D$. Therefore, we conclude that $\cos{\alpha}=\dfrac{1}{\sqrt{10}}$. Hence $AC=BC=\sqrt{\dfrac{5}{2}}$. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,P,K; O=(0,0);A=dir(216.86989792);C=dir(0);B=dir(143.13010208);D=foot(B,A,C);P=incenter(B,C,D);K=midpoint(A--C); draw(A--B--C--cycle,org);draw(B--K,org);draw(B--D,org); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$P$",P,dir(P)); dot("$K$",K,dir(K)); [/asy][/asy]