Problem Analysis: We are given a circle with diameter PQ and center O. A point R is on the circle such that $\angle RPO = 39^\circ$. We need to find the measure of $\angle RQP$. We will use properties of isosceles triangles and the angle subtended by a diameter.

Diameter Property in Circle

Since PQ is the diameter of the circle, the angle subtended by the diameter at any point on the circumference is $90^\circ$. Therefore, $\angle PRQ = 90^\circ$.

Isosceles Triangle RPO

Consider the triangle $\triangle RPO$. The sides OP and OR are both radii of the same circle. Hence, $OP = OR$. This makes $\triangle RPO$ an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal. Given $\angle RPO = 39^\circ$, the angle opposite to OP, which is $\angle ORP$, must also be equal to $\angle RPO$.

So, $\angle ORP = \angle RPO = 39^\circ$.

Calculate Angle ORQ

We know that $\angle PRQ = \angle ORP + \angle ORQ$.

Substituting the known values:

$ 90^\circ = 39^\circ + \angle ORQ $

Solving for $\angle ORQ$: $ \angle ORQ = 90^\circ - 39^\circ $
$ \angle ORQ = 51^\circ $

Isosceles Triangle RQO

Now consider the triangle $\triangle RQO$. The sides OQ and OR are both radii of the circle. Hence, $OQ = OR$. This makes $\triangle RQO$ an isosceles triangle.

The angles opposite the equal sides are equal. Therefore, the angle opposite to OR ($\angle OQR$, which is the same as $\angle RQP$) must be equal to the angle opposite to OQ ($\angle ORQ$).

So, $\angle RQP = \angle ORQ$.

Since we calculated $\angle ORQ = 51^\circ$, we have:

$ \angle RQP = 51^\circ $