Question

If in a \(\Delta ABC\), \(AB = 6\) cm and \(DE \parallel BC\) such that \(AE = \frac{1}{3} AC\), then the length of BD is

• A. 2 cm
• B. 3 cm
• C. 4 cm
• D. 5 cm

Hint:

An alternative way to use BPT directly:
If \(AE = \frac{1}{3} AC\), then \(EC = \frac{2}{3} AC\).
Thus, \(\frac{AD}{DB} = \frac{AE}{EC} = \frac{1}{2} \implies DB = 2 AD\).
Since \(AD + DB = 6 \implies AD + 2AD = 6 \implies AD = 2 \text{ cm}, DB = 4 \text{ cm}\).

Answer 1

The Correct Option is C

Step 1: Convert the given ratio to a side-split ratio.
Since $AE = \frac{1}{3}AC$, we get $EC = \frac{2}{3}AC$, so $AE:EC = 1:2$.
Step 2: Apply the Basic Proportionality Theorem directly on AD and DB.
Because $DE \parallel BC$, $\frac{AD}{DB} = \frac{AE}{EC} = \frac{1}{2}$, meaning $DB = 2\,AD$.
Step 3: Use the total length of AB.
\[ AD + DB = 6 \implies AD + 2AD = 6 \implies AD = 2 \]
So $DB = 2 \times 2$.
\[ \boxed{4 \text{ cm}} \]