Help me find a side of a quadrilateral given one side and distance between the centers of two circles inscribed in the same quadrilateral

Question

For the quadrilateral ABCD it's given that AD=2, $\angle$ABD=$\angle$ACD=90$^\circ$. The distance between the centers of the inscribed circles in $_\triangle$ABD and $_\triangle$ACD is $\sqrt{2}$. Find the length of BC.

Answer 1

Interesting question.

The constraints $\widehat{ABD}=\widehat{ACD}=90^\circ$ give that both $B$ and $C$ lie on the circle with diameter $AD$. Assuming that both $B$ and $C$ lie on the same side of $AD$, it follows (by angle chasing) that both the incenter $I_B$ of $ABD$ and the incenter $I_C$ of $ACD$ lie on the arc of circle through $A$ and $D$ having center at $P$ (defined as the intersection of the perpendicular bisector of $AD$ with the circumcircle of $ABD$, in such a way that $P$ and $B$ lie on opposite sides of $AD$). Since $AP=\sqrt{2}$, the constraint $I_B I_C=\sqrt{2}$ implies that $\widehat{I_B P I_C}=60^\circ$. On the other hand both $BI_B$ and $CI_C$ go through $P$, hence $\widehat{BPC}=60^\circ$ and the length of $BC$ is given by $\sqrt{3}$ times the circumradius of $ABD$.

$$ \boxed{BC=\color{red}{\sqrt{3}}}. $$