The question is in the title, and I need a hint or a solution:
A convex quadrilateral is separated by its diagonals into 4 triangles. Suppose these triangles have congruent incircles, then show that the quadrilateral must be a rhombus.
I can see that the quadrilateral whose vertices are incentres (of the 4 triangles described in the problem) is a rhombus. Looks like I have to show that this rhombus is a square. But I am not able to prove it.
I would also appreciate a reference which deals with problems of this type. I found a reference with similar problems and would love to see more.
Hint: Flip the construction in the question around. Instead of starting with the convex quadrilateral to get the congruent circles, we start with the congruent circles and hope to get a quadrilateral.
Consider a configuration where we have 2 lines that intersect at $X$, and we place 4 congruent circles that are tangential to both lines.
Suppose we pick a point $A$ on a line, and draw a tangent to the circle to reach point $B$, draw a tangent to the next circle to reach point $C$, draw a tangent to the next circle to reach point $D$, draw a tangent to the next circle to reach point $E$.
Show that $ A = E$ (meaning we have a quadrilateral) iff the lines are perpendicular (and hence show that they bisect each other).
Corollary: "The incircles are congruent" iff "the diagonals are perpendicular and bisect each other" iff "the quadrilateral is a rhombus".