# An angular inequality

In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $$\varphi \le \max \{ \angle BAC, \angle ABC \}$$ This is a problem from the IMO Shortlist. I have no idea about it. Thanks for helping.

• If it's from the IMO shortlist, it should be easy to search for a solution, since those are released. If it's from IMO SL 2014, you should not be posting it. – Calvin Lin Jul 12 '14 at 15:00
• It is from 2013. I don't have any solution @CalvinLin so I posted it, I know also that we shouldn't post it on the web. :) – shadow10 Jul 12 '14 at 15:18
• Ah, from 2013 it would be harder to find a solution. Wait a few days and someone will post the pdf somewhere lol. – Calvin Lin Jul 12 '14 at 15:19

Just a first step, i.e. a way to inscribe a rhombus in a convex quadrilateral $ABCD$. Fix the direction of the diagonals of the rhombus, let them be parallel to lines $r$ and $s$, with $r\perp s$. Consider the segments parallel to $r$ having their extrema on $AB$ and $CD$: the center of the rhombus must belong to the locus of midpoints of such segments, hence it must lie on a certain line $t$. If we consider the segments parallel to $s$ having their extrema on $BC$ and $AD$, the center of the rhombus must belong to the locus of midpoints of such segments, too, hence it must lie on a certain line $u$, and the center of the rhombus is just $R=t\cap u$ (hence it lies on a hyperbola). Once we have the center, we can find the vertices of the rhombus by intersecting the lines parallel to $r,s$ through $R$ with the sides of $ABCD$.