Question

In the triangle ABC the angles bisectors are drawn from vertices A and B intersect at the point D. Find the measure of angle ABD if a measure of angle A =a, measure of B=b

Answer 1

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 \( \angle B \):

1. The angle bisector of \( \angle A \) splits it into two equal parts:
\[
\angle ABD = \frac{1}{2} a
\]

2. 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:
\[
\text{The measure of angle } ABD \text{ is } \frac{b}{2}.
\]