Question

In \(\triangle ABC\), D lies on AB and E lies on AC such that AD = 2 cm, DB = 3 cm, AE = 4 cm and EC = 6 cm. By the converse of the Basic Proportionality Theorem, ________.

• A. DE \(\perp\) BC
• B. DE \(\parallel\) AB
• C. DE bisects BC
• D. DE \(\parallel\) BC

Hint:

Whenever you are given BPT-like segment values, quickly compute and compare the two ratios.
If they are equal, the lines are parallel.
Here, \(2/3\) is obviously equal to \(4/6\), leading directly to \(DE \parallel BC\).

Answer 1

The Correct Option is D

Step 1: Work out each side's split ratio.
On side $AB$, the ratio is \[ \frac{AD}{DB} = \frac{2}{3} \] On side $AC$, the ratio is \[ \frac{AE}{EC} = \frac{4}{6} = \frac{2}{3} \]
Step 2: Compare the two ratios.
Both ratios simplify to the same value, $\frac{2}{3}$, so $DE$ divides $AB$ and $AC$ in exactly the same proportion.
Step 3: Recall what the converse of BPT says.
If a line divides two sides of a triangle in the same ratio, it must be parallel to the third side.
Step 4: Apply it here.
Since the ratios match, line $DE$ must be parallel to $BC$.
\[ \boxed{DE \parallel BC} \]