In triangle ABC, angle ABC is 15 degrees. Point D lies on side BC such that AD = BD. What is the measure of angle ADC (in degrees)?

Introduction / Context:
This is a geometry problem involving special angle constructions in a triangle. Triangle ABC has angle ABC equal to 15 degrees, and a point D is chosen on side BC such that AD equals BD. We are asked to find angle ADC. This is a nontrivial configuration and requires careful geometric reasoning or construction, often using properties of isosceles triangles and angle chasing. Such questions appear in higher level aptitude and Olympiad style geometry sections.

Given Data / Assumptions:
• Triangle ABC is any triangle with angle ABC = 15 degrees.
• Point D lies on side BC of triangle ABC.
• Length AD equals length BD (AD = BD).
• We must determine the measure of angle ADC in degrees.
• The configuration implicitly determines D uniquely so that AD = BD is satisfied.

Concept / Approach:
Because AD = BD, triangle ABD is isosceles with AB as the base. The given angle at B in triangle ABC is small (15 degrees), which suggests that known angle splitting patterns like 15, 30, 45, and 60 degrees may arise. A common approach is to place the triangle on coordinates or use classical angle chasing. If we place B at the origin, take BC along the x axis, and choose A at an appropriate angle, we can solve using analytic geometry, enforcing AD = BD to determine D, and then compute angle ADC. It turns out that this configuration leads to angle ADC equal to 30 degrees.

Step-by-Step Solution:
Step 1: Conceptually, let us place B at (0, 0) and C along the x axis. Let BC = 1 unit for convenience, so C is at (1, 0). Step 2: Angle ABC is 15 degrees, so side BA makes an angle of 15 degrees with BC. Take AB = 1 unit for convenience, so A has coordinates (cos 15°, sin 15°). Step 3: Let D be at (t, 0) along BC, with 0 < t < 1. Then BD = t and AD is the distance between A and D. Step 4: Impose the condition AD = BD. Squaring both distances gives an equation in t. Solving this yields t ≈ 0.5176, which lies between 0 and 1, so D is indeed on BC. Step 5: With the coordinates of A, D, and C known, compute the angle at D formed by segments DA and DC using the vector or dot product method. The resulting angle ADC is found to be exactly 30 degrees.

Verification / Alternative check:
Instead of coordinates, one can use synthetic geometry. Since AD = BD, triangle ABD is isosceles with angles at A and B equal. If angle B in triangle ABC is 15 degrees, then angle ABD equals angle BAD = 15 degrees. With appropriate constructions and focusing on triangle ADC, angle chasing using known sums of angles in triangles can show that angle ADC must be 30 degrees. The coordinate method and synthetic approach both agree on 30 degrees, so the result is confirmed.

Why Other Options Are Wrong:
15 degrees would make angle ADC equal to angle ABC, which contradicts the internal isosceles structure created by AD = BD.
45 and 60 degrees do not fit the precise angle relations that come from AD = BD and ABC = 15 degrees when checked by detailed angle chasing or coordinate geometry.
75 degrees is too large given the geometry and does not arise from any consistent configuration respecting all conditions. Only 30 degrees is compatible with both the distance equality and the angle structure.

Common Pitfalls:
A major pitfall is assuming D is some simple point like the midpoint of BC without verifying the AD = BD condition. Another error is applying angle bisector theorems where they do not apply, or assuming AD bisects some angle. Mismanaging the trigonometric or coordinate computations can also lead to incorrect numerical values. Drawing an accurate diagram, using either coordinate geometry or systematic angle chasing, and carefully enforcing AD = BD are key to solving this problem correctly.

Final Answer:
The measure of angle ADC is 30 degrees.