If $\frac{AB}{AC} = \frac{BD}{DC}$ then $\angle ABC$ is:
To solve this problem, we need to use the Angle Bisector Theorem. According to this theorem, if a line divides the opposite side of a triangle in the ratio of the other two sides, then it is an angle bisector of the opposite angle.
In the given triangle, we have:
\(\frac{AB}{AC} = \frac{BD}{DC}\)
This means that \(AD\) is the angle bisector of \(\angle BAC\). Therefore, \(\angle BAD = \angle DAC\).
Given \(\angle BAD = 28^\circ\) and \(\angle DAC = 28^\circ\), the total angle at \(A\) is:
\(\angle BAC = \angle BAD + \angle DAC = 28^\circ + 28^\circ = 56^\circ\)
We also know that \(\angle ACD = 60^\circ\).
Since \(\angle ACD\) and \(\angle ABC\) are angles of triangle \(ABC\), we can use the triangle angle sum property:
\(\angle ABC + \angle BAC + \angle ACD = 180^\circ\)
Substituting the known values, we get:
\(\angle ABC + 56^\circ + 60^\circ = 180^\circ\)
Solving for \(\angle ABC\):
\(\angle ABC = 180^\circ - 56^\circ - 60^\circ = 64^\circ\)
Therefore, the correct answer is
$64^\circ$
.
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