A rectangular strip of paper edge $AC$ is first folded in half making fold crease through mid-point $M$ and again bent folded along a new line through $C$ adjusted such that corner $A$ falls on divider $MB$. Show that triangle $ABC$ is equilateral.
EDIT1:
Rajendra Kumar makes a full Icosahedron repeated in this way!
This is an interesting problem. Thank you for posting it.
Construct $BD$ perpendicular to one edge of the paper $$ BD = MC = \frac{1}{2}AC = \frac{1}{2}BC$$ It follows that $$\angle BCD = 30^\circ\\ \angle MCB = 60^\circ $$ From the condition that you give, $$AC = BC$$ Therefore $\Delta ACB$ is an equilateral triangle.