Inscribed and Tangent

Let’s designate $\angle BCG=\alpha$ . Radius of the inscribed circle in triangle $BCD$ is its area divided by a semi-perimeter. $R=\frac{sin(\alpha)cos(\alpha)}{1+sin(\alpha)}$ .

Also $R+x=\frac{1}{2}$ and $\frac{R}{x}=tan(\alpha$ ). So we have $\frac{R}{tan(\alpha)}+R=\frac{1}{2}$ . This will give us $R=\frac{sin(\alpha)}{2(sin(\alpha)+cos(\alpha))}$

Setting the two expressions for $R$ equal to each other, we get $\frac{cos(\alpha)}{1+sin(\alpha)}=\frac{1}{2(sin(\alpha)+cos(\alpha))}$ .

Introducing $y=sin(\alpha)$ we can write this equation as $\frac{\sqrt{1-y^2}}{1+y}=\frac{1}{2(y+\sqrt{1-y^2})}$ .

The applicable solution is $y=0.77569$ , giving us $\alpha=50.868^\circ$

The $\angle ABC=\angle BCD=2\times\alpha=101.736^\circ$ .