Question

theorem: a line parallel to one side of a triangle divides the other two proportionately.
in the figure below, segment de is parallel to segment bc and segment ef is parallel to ab:
which statement can be proved true using the given theorem?
segment bf = 16
segment bd = 20
segment bd = 15
segment bf = 32

Explanation:

Step1: Apply the theorem to \(\triangle ABC\) with \(EF\parallel AB\)

By the theorem (a line parallel to one side of a triangle divides the other two proportionately), in \(\triangle ABC\) since \(EF\parallel AB\), we have \(\frac{CE}{EA}=\frac{CF}{FB}\). Given \(CE = 20\), \(EA=24\), \(CF = 30\). Let \(BF=x\). Then \(\frac{20}{24}=\frac{30}{x}\). Cross - multiply: \(20x=24\times30\), so \(x=\frac{24\times30}{20}=36\). This approach is wrong. Let's use the theorem for \(DE\parallel BC\) in \(\triangle ABC\).

Step2: Apply the theorem to \(\triangle ABC\) with \(DE\parallel BC\)

In \(\triangle ABC\), since \(DE\parallel BC\), we have \(\frac{AD}{DB}=\frac{AE}{EC}\). Let \(BD = x\). Then \(\frac{18}{x}=\frac{24}{20}\). Cross - multiply: \(24x=18\times20\).

Step3: Solve for \(x\)

\(x=\frac{18\times20}{24}=\frac{360}{24}=15\)

Answer:

Segment \(BD = 15\)