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.
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$\begingroup$ I would think the quadrilateral should be $ABDC$ or $ACBD$ but $ABCD$ is not possible, as it seems that $AD$ is a diagonal, not a side. $\endgroup$gt6989b– gt6989b2017-05-17 19:24:20 +00:00Commented May 17, 2017 at 19:24
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$\begingroup$ Well in this problem AD is actually a side of the quadrilateral. It's easy to prove that AD is the diameter of the circumscribed circle of ABCD. I am stuck with the further solution. $\endgroup$Ivalin– Ivalin2017-05-17 19:48:34 +00:00Commented May 17, 2017 at 19:48
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$\begingroup$ @gt6989b: there is nothing wrong with the given constraints, have a look at the picture below (with $C\equiv C_2$). $\endgroup$Jack D'Aurizio– Jack D'Aurizio2017-05-17 20:04:46 +00:00Commented May 17, 2017 at 20:04
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$\begingroup$ @JackD'Aurizio I see, i interpreted both given angles to be actual angles of the quadrilateral... $\endgroup$gt6989b– gt6989b2017-05-17 20:47:24 +00:00Commented May 17, 2017 at 20:47
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1 Answer
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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 $ABD$.
$$ \boxed{BC=\color{red}{\sqrt{3}}}. $$
