In triangle `ABC, /_ABC = 15^@`. D is a point on BC such that `AD = BD`. What is the measure of `/_ADC `(in degrees)?

Options

A. 15
B. 30
C. 45
D. 60

Answer

To solve the problem, we need to find the measure of angle \( \angle ADC \) in triangle \( ABC \) where \( \angle ABC = 15^\circ \) and point \( D \) is on line segment \( BC \) such that \( AD = BD \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Triangle \( ABC \) has \( \angle ABC = 15^\circ \).
- Point \( D \) is on segment \( BC \) such that \( AD = BD \).

2. **Understand the Implications of \( AD = BD \)**:
- Since \( AD = BD \), triangle \( ABD \) is an isosceles triangle with \( AB = AD \) and \( BD = AD \).
- This means that the angles opposite to these equal sides are also equal. Therefore, \( \angle ADB = \angle ABD \).

3. **Calculate \( \angle ABD \)**:
- Let \( \angle ABD = x \).
- In triangle \( ABC \), the sum of angles is \( 180^\circ \). Therefore, we can write:
\[ \angle ABC + \angle ACB + \angle CAB = 180^\circ \]
Since \( \angle ABC = 15^\circ \), we have:
\[ 15^\circ + \angle ACB + \angle CAB = 180^\circ \]
This simplifies to:
\[ \angle ACB + \angle CAB = 165^\circ \]

4. **Use the Isosceles Triangle Property**:
- In triangle \( ABD \):
\[ \angle ADB + 2x = 180^\circ \]
- Since \( \angle ADB = 180^\circ - 15^\circ - x \) (because \( \angle ADB + \angle ABC + \angle ABD = 180^\circ \)), we can substitute:
\[ 180^\circ - 15^\circ - x + 2x = 180^\circ \]
Simplifying gives:
\[ 165^\circ + x = 180^\circ \]
Thus, \( x = 15^\circ \).

5. **Find \( \angle ADC \)**:
- Since \( D \) is on \( BC \) and \( AD = BD \), triangle \( ADC \) will have:
\[ \angle ADC = \angle ABD = 15^\circ \]

### Conclusion:
The measure of \( \angle ADC \) is \( 15^\circ \).