Darts and Kites

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This solution is from Arun Iyer of SIA High School and Junior College.

Part 1

First I will show that $POR$ is a straight line. For this I would like to state the perpendicular bisector theorem.

PERPENDICULAR BISECTOR THEOREM: Every point equidistant from the two ends of a line segment lies on the perpendicular bisector of the line segment.

Now consider the line segment $QS$.

$OQ=OS=1$ therefore by the perpendicular bisector theorem, $O$ must lie on the perpendicular bisector of $QS$.

$PQ=PS$ (as the sides of the rhombus are equal), therefore by the perpendicular bisector theorem, $P$ must lie on the perpendicular bisector of $QS$.

$RQ=RS$ (as the sides of the rhombus are equal), therefore by the perpendicular bisector theorem, $R$ must lie on the perpendicular bisector of $QS$.

Now the perpendicular bisector of a line segment is unique and hence $P$, $O$, $R$ must lie on the same perpendicular bisector and hence $POR$ is a straight line.

Part 2

Now I will get all the angles of the rhombus.

$\angle QPS$ = $72^{\circ}$ (given), $\angle QRS = \angle QPS = 72^{\circ}$ as they are opposite angles of a rhombus. The diagonal of the rhombus bisects the angles of a rhombus and therefore $\angle QPO = \angle SPO = \angle QRO = \angle SRO = 36^{\circ}$.

Triangles $OQR$ and $OSR$ are isosceles triangles, therefore $\angle OSR = \angle OQR = 36^{\circ}$.

Using the fact that sum of angles of a triangle is 180 degrees, we can see that $\angle SOR$ and $\angle QOR$ are equal to 108 degrees. Since we have proved that $POR$ is a straight line in Part 1, we can determine $\angle QOP$ and $\angle SOP$ to be 72 degrees.

Again using the fact that sum of angles of a triangle is 180 degrees, we can see that $\angle OQP$ and $\angle OSP$ are equal to 72 degrees.

Part 3