Question

3.11.3 more proportional relationships?
de is parallel to ac.
find the lengths of ac and ad.
side | length | correct?
ac | |
ad | |
(image: a geometric figure with points a, d, b, e, c. bd is length 2, be is length 4, ec is length 6, de is length 5. de is parallel to ac.)

Explanation:

Step1: Use the basic proportionality theorem (Thales' theorem)

Since \(DE\parallel AC\), we have \(\frac{BD}{BA}=\frac{BE}{BC}\). Let \(AD = x\), then \(BA=BD + AD=2 + x\). And \(BE = 4\), \(BC=BE + EC=4 + 6=10\). But also, from the similar - triangles ( \(\triangle BDE\sim\triangle BAC\) because \(DE\parallel AC\)), we can use the ratio of corresponding sides. The ratio of sides in \(\triangle BDE\) and \(\triangle BAC\) is \(\frac{BD}{BA}=\frac{BE}{BC}=\frac{DE}{AC}\).

Another way: Using the property of similar triangles \(\triangle BDE\sim\triangle BAC\). The ratio of sides is \(\frac{BD}{BA}=\frac{BE}{BC}\). Let's first find \(AD\).
We know that \(\frac{BD}{BA}=\frac{BE}{BC}\). Substituting \(BD = 2\), \(BE = 4\), \(BC=4 + 6=10\). Let \(BA=BD + AD=2+AD\). So \(\frac{2}{2 + AD}=\frac{4}{10}\).
Cross - multiply: \(4(2 + AD)=2\times10\).

Step2: Solve for \(AD\)

Expand \(4(2 + AD)=2\times10\) gives \(8+4AD = 20\).
Subtract 8 from both sides: \(4AD=20 - 8=12\).
Divide both sides by 4: \(AD=\frac{12}{4}=3\).

Step3: Solve for \(AC\) using the ratio \(\frac{DE}{AC}=\frac{BE}{BC}\)

We know \(DE = 5\), \(BE = 4\), \(BC = 10\). Let \(AC=y\). Then \(\frac{5}{y}=\frac{4}{10}\).
Cross - multiply: \(4y=5\times10\).
\(4y = 50\), so \(y=\frac{50}{4}=12.5\).

Answer:

\(AC = 12.5\), \(AD = 3\)