Let $r$ denote radius of incircle of $\triangle{ABC}$, we have $ID=IE=\sqrt{r^2+\Big( \frac{a+b+c}{2}\Big)^2}$. By law of cosine, $DE^2=b^2+(a+c)^2+2\times b\times (a+c)\times \frac{a^2+c^2-b^2}{2ac}$. We want to show that $\angle{DIE}=180^{\circ}-\angle{ABC}$. By law of cosine, it's equivalent to $DE^2=2\Big( r^2+\Big( \frac{a+b+c}{2}\Big)^2\Big) +2\times \Big( r^2+\Big( \frac{a+b+c}{2}\Big)^2\Big) \times \frac{a^2+c^2-b^2}{2ac}$. $\Leftrightarrow r^2\Big( 2+\frac{a^2+c^2-b^2}{2ac}\Big) =b^2+(a+c)^2+\Big( b(a+c)-\Big( \frac{a+b+c}{2}\Big)^2 \Big) \frac{a^2+c^2-b^2}{2ac} -\Big( \frac{a+b+c}{2}\Big)^2$ $\Leftrightarrow r^2\frac{(a+c-b)(a+b+c)}{ac}=\frac{(b+c-a)(a+b-c)(a+c-b)^2}{4ac}$ $\Leftrightarrow r^2=\frac{4A^2}{a+b+c}$ where $A$ denote area of $\triangle{ABC}$ which is true.

Let $XYZ$ be intouch triangle of $\triangle ABC$($X\in AB , Y\in BC , Z\in AC$).We can see $$DX=DA+AX=BC+AZ=BY+CY+AZ=BY+CZ+AZ=BY+AC=BY+BE=EY $$and $IX=IY=r $ ($r$ is inradius of $\triangle ABC$).Now we get $\triangle IXD \cong \triangle IYE \Longleftrightarrow \angle{IDX}=\angle{IEY} \Longleftrightarrow D,E,B,I$ are concyclic and the proof is complete.

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Let $D^*$ a touch point with $BC$ it's easy to see that $BD-BD^*=ED^*$ the diagram clarify the remaining of the proof : RH HAS

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