4. Given a rhombus $ABCD$ with $\measuredangle BAD=60^{\circ}$. The angle bisectors of the angles between the diagonals intersect the sides of the rhombus at points $M, N, P$ and $Q$.

a) What type of quadrilateral is quadrilateral $MNPQ$?

b) If point $M$ lies on side $AB$, calculate the ratio $AM: MB$.

c) Determine the ratio of those segments of the larger and smaller diagonals of the rhombus that lie outside quadrilateral $MNPQ$.
Solution. a) From the congruence of triangles $O B M$ and $O B N$ it follows that $O M=O N$, and from the congruence of triangles $O A M$ and $O A Q$ it follows that $O M=O Q$ (see diagram on the right). Similarly, $O Q=O P$. From this, we conclude that $M P=N Q$ and since $\measuredangle M O N$ is a right angle, it follows that quadrilateral $M N P Q$ is a square.

b) Triangle $A M Q$ is equilateral. Let the side of this triangle be denoted by $x$. Then triangle $B R M$, which is half of an equilateral triangle, has a height $M R=\frac{x}{2}$. Therefore, $M R=M B \frac{\sqrt{3}}{2}$, and if we set $M B=y$, we get $\frac{x}{2}=y \frac{\sqrt{3}}{2}$, i.e., $x=y \sqrt{3}$. Thus, $A M: M B=x: y=y \sqrt{3}: y=\sqrt{3}: 1$.

c) From triangle $A M Q$ it follows that $A S=\frac{x \sqrt{3}}{2}=y \sqrt{3} \frac{\sqrt{3}}{2}=\frac{3 y}{2}$, and from triangle $B R M$ it follows that $B R=\frac{y}{2}$. Therefore, $A S: B R=\frac{3 y}{2}: \frac{y}{2}=3: 1$.