Question

segment \\(\overline{bd}\\) bisects \\(\angle abc\\). solve for \\(x\\). round to the nearest tenth, if necessary. (image not necessarily to scale.)

Explanation:

Step1: Apply the Angle - Bisector Theorem

The Angle - Bisector Theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides.
For \(\triangle ABC\) with \(BD\) bisecting \(\angle ABC\), we have \(\frac{AD}{DC}=\frac{AB}{BC}\).
Let \(AD = x\), \(DC = 6\), \(AB = 15\), and \(BC = 20\).
Substituting the values into the proportion \(\frac{x}{6}=\frac{15}{20}\).

Step2: Solve the proportion for \(x\)

Cross - multiply: \(20x=15\times6\).
Since \(15\times6 = 90\), the equation becomes \(20x = 90\).
Divide both sides by 20: \(x=\frac{90}{20}\).
Simplify \(\frac{90}{20}=\frac{9}{2}=4.5\).

Answer:

\(x = 4.5\)