In an isosceles triangle ABC, sides AB and AC are equal and BC is the unequal base. AD is the median to side BC, meeting BC at D. Ray DP is the angle bisector of angle ∠ADB and PQ is drawn parallel to BC, meeting AC at Q. What is the measure of angle ∠PDQ in degrees?

Introduction / Context:
This is a more advanced geometry question involving an isosceles triangle, a median to the base, an angle bisector, and a line parallel to the base. It tests deep understanding of symmetry and angle relationships inside triangles, and is typical of higher level aptitude and Olympiad style geometry problems.

Given Data / Assumptions:
- Triangle ABC is isosceles with AB = AC and base BC unequal to these sides.
- AD is the median to BC, so D is the midpoint of BC.
- DP is the angle bisector of angle ∠ADB, so P lies on side AB and DP divides ∠ADB into two equal angles.
- PQ is drawn parallel to BC and meets AC at Q.
- We must find the measure of angle ∠PDQ in degrees.

Concept / Approach:
Because AB = AC and AD is the median to the base BC, in an isosceles triangle this median is also an altitude and an angle bisector of the vertex angle at A. Thus AD is a line of symmetry that splits the triangle into two congruent right triangles. The constructions involving the angle bisector at D and the line parallel to BC create a symmetric configuration around AD that leads to right angles at D inside the smaller quadrilateral formed by P, D, and Q.

Step-by-Step Solution:
Step 1: Note that in isosceles triangle ABC with AB = AC and median AD to base BC, AD is perpendicular to BC and also bisects angle ∠BAC. Step 2: Thus triangles ABD and ACD are congruent right triangles with AD as common altitude and BD = DC. Step 3: At point D on BC, consider angle ∠ADB formed by segments DA and DB. Ray DP is constructed as the angle bisector of ∠ADB, so it lies symmetrically between DA and DB. Step 4: Because of the overall symmetry of the figure, a corresponding construction on the other side with line through P parallel to BC meeting AC at Q ensures that D is the midpoint of segment PQ and that segment PQ is parallel to BC. Step 5: When this is worked out carefully using coordinate geometry or reflections, one finds that P and Q are symmetric with respect to the vertical line AD, and that the quadrilateral PDQ is arranged so that angle ∠PDQ is a right angle. Step 6: Detailed coordinate calculations (for example, by placing B at (−1, 0), C at (1, 0), and A at (0, h)) confirm that P and Q have equal y coordinates and opposite x coordinates, and D lies at the origin, making ∠PDQ = 90°.

Verification / Alternative Check:
Using a specific numerical example, choose coordinates B(−1, 0), C(1, 0), and A(0, 5). Then D is (0, 0) as the midpoint of BC. By constructing the angle bisector at D and the parallel through P to BC, one obtains P(−5/6, 5/6) and Q(5/6, 5/6). Vectors DP and DQ are then perpendicular, since their dot product is zero, proving that ∠PDQ = 90°. This coordinate verification is consistent across any choice of height h, so the right angle is a geometric consequence of the given configuration, not an artifact of a particular drawing.

Why Other Options Are Wrong:
Option a, 130°, and option d, 45°, do not match the orthogonal arrangement of DP and DQ that arises from the symmetry of the isosceles triangle and the parallel line construction.
Option c, 180°, would imply that P, D, and Q are collinear, which contradicts the fact that PQ is parallel to BC and passes above the base, creating a genuine interior angle at D rather than a straight angle.

Common Pitfalls:
Because the configuration is complex, it is easy to misplace points P and Q or misinterpret which segments are parallel. Some students attempt to guess an angle based on rough sketches, which can be misleading. Others forget that in an isosceles triangle the median to the base is also an altitude and angle bisector, which is a powerful symmetry property. A careful construction or a coordinate geometry approach, treating AD as an axis of symmetry, provides a clear path to the correct right angle at D.

Final Answer:
The measure of angle ∠PDQ is 90°.