To solve for the measure of angle \( ABD \) in triangle \( ABC \) with angle measures \( \angle A = a \) and \( \angle B = b \), we will utilize some properties of angle bisectors.
Recall that angle \( D \) is formed at the intersection of the angle bisectors of \( \angle A \) and \( \angle B \):
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The angle bisector of \( \angle A \) splits it into two equal parts: \[ \angle ABD = \frac{1}{2} a \]
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The angle bisector of \( \angle B \) splits it into two equal parts: \[ \angle DBC = \frac{1}{2} b \]
Now, we can find the measure of angle \( ADB \), which is the supplement of angle \( DAB + ABD \): \[ \angle ADB = 180^\circ - \left(\frac{1}{2} a + \frac{1}{2} b\right) \] This simplifies to: \[ \angle ADB = 180^\circ - \frac{1}{2}(a + b) \]
To find \( \angle ABD \): \[ \angle ABC = 180^\circ - (a + b) \quad \text{(since the sum of angles in triangle is } 180^\circ) \]
Using these angles, we can set up the measures: \[ ABD + ABC + ADB = 180^\circ \]
Substituting the values we derived: \[ \frac{1}{2} a + ABD + \left(180^\circ - \frac{1}{2}(a + b) \right) = 180^\circ \]
We can solve for the angle measure of \( ABD \): \[ ABD = 180^\circ - \frac{1}{2} a - \left(180^\circ - \frac{1}{2}(a + b) \right) \]
This can be simplified into: \[ ABD = \frac{1}{2}(b) \]
Thus, the measure of angle \( ABD \) is: \[ \angle ABD = \frac{b}{2} \]
In conclusion: \[ \text{The measure of angle } ABD \text{ is } \frac{b}{2}. \]