Show the congruence $YZ+JK=JF$ in this square with trapezoids, triangles constructed with square's midpoints

Question

Show that $YZ+JK=JF$

I tried Pythagorean theorem.

OG: Area of a square inside a square created by connecting point-opposite midpoint

Answer 1

Let $\angle{ACB} = \theta$. Now apply $\tan\theta$, it gives $\frac12$. It is same with angles $DFC, KLF, LAY,$ so these angles are equal.

Hence $\triangle{AXB}, \triangle{LYZ}, \triangle{FJK}$ and $\triangle{CED}$ are congruent by ASA rule.

We can also see some similar triangles. These are $\triangle{LYZ} $ is similar to $\triangle{LJF}$, and etc. So $LY = YJ$ and etc. Also $2YZ = JF$ and $2JK = CE$. But $CE = JF$, so $YZ = JK = \frac{1}{2}JF$.

Hence $YZ + JK = JF$ (Hence Proved)