Question

03- proportions in similar triangles
when we see a picture like the one below, it isnt as easy to see the proportion because the
triangles are overlapping. draw 2 separate triangles first every time!!

1. in right triangle abc shown below, point d is on \\(\overline{ab}\\) and point e is on \\(\overline{cb}\\) such that \\(\overline{ac} \parallel \overline{de}\\).

if \\(ab = 18\\), \\(bc = 20\\), and \\(ad = 6\\), what is the length of \\(be\\),
to the nearest tenth?

1. using the diagram to the right, find the value

of \\(x\\)

Explanation:

Step1: Identify similar triangles

Given right triangle \(ABC\) with \(AC \parallel DE\), \(\triangle ABC \sim \triangle DBE\) (AA similarity: shared \(\angle B\), both right angles).

Step2: Set up proportion

\(\frac{AD}{AB} = \frac{CE}{CB}\). Let \(CE = x\), then \(BE = 10 - x\). Wait, correct proportion: \(\frac{AD}{AB} = \frac{CE}{CB}\)? No, \(\frac{AD}{AB} = \frac{CE}{CB}\) is wrong—should be \(\frac{DB}{AB} = \frac{BE}{BC}\). \(DB = AB - AD = 15 - 6 = 9\). So \(\frac{9}{15} = \frac{BE}{10}\).

Step3: Solve for BE

\(BE = \frac{9 \times 10}{15} = 6\).

Step4: Solve for x (second problem)

Triangle with parallel line: \(\frac{2}{2+4} = \frac{x}{7}\) → \(\frac{2}{6} = \frac{x}{7}\) → \(x = \frac{2 \times 7}{6} = \frac{7}{3} \approx 2.3\).

Answer:

1. 6
2. \(\frac{7}{3}\) (or ~2.3)