Finding Side length of Rhombus inside Isosceles Triangle.

Hi, after trying to figure out this problem for a few hours i think i need some guidance.

The question is given AB=BC=15, and AC=18, and ADEF is a rhombus.

How do i find 1.) the side of the rhombus and 2.) the radius of a inscribed circle of the rhombus?

Here is a picture the teacher gave us, not drawn to scale.
Attachment 21730

I know triangle ABC is isosceles , and triangle DBE and FEC are congruent and my intuition tells me the side length of the rhombus is 9, but i don't know how to prove it.

Hello aquameatwad,
Draw a sketch similar to that given but add a perpendicular bisector of AC meeting DE @ G and AC @H.Calculate the length of BH using Pythagoros.Using the angle bisector theorem calculate the lengths ofBE and EC.
Using similar triangles find length of BG,Using Pythagoros find GE.Side of rhombus side = 2*GE




bjh

Let side of rhombus is x. Then for similar triangles BED and ECF

CF/ED = EF/BD => (18 - x )/x = x/( 15 - x )

solving equation we get x = 90/11 but not 9