Question

In \( \triangle ABC \), \( DE \parallel BC \), \( \frac{AD}{DB} = \frac{2}{3} \) and \( AC = 2.5 \text{ cm} \), then \( AE = \_\_\_\_\_ \text{ cm}\).

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Whenever ratios like \(2:3\) are given, convert them into parts immediately. Here total parts become \(2+3=5\), making proportional calculations much easier.

Answer 1

The Correct Option is D

Concept: When a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. This is called the Basic Proportionality Theorem (BPT) or Thales' Theorem. If: \[ DE \parallel BC \] then: \[ \frac{AD}{DB} = \frac{AE}{EC} \] Also, another useful proportional form is: \[ \frac{AD}{AB} = \frac{AE}{AC} \] This form becomes easier when the entire side length is given.

Step 1: Understand the given ratio. We are given: \[ \frac{AD}{DB} = \frac{2}{3} \] This means side \(AB\) is divided into two parts in the ratio \(2:3\). So we may assume: \[ AD = 2k,\quad DB = 3k \] Therefore: \[ AB = AD + DB = 2k + 3k = 5k \] Thus: \[ \frac{AD}{AB} = \frac{2k}{5k} = \frac{2}{5} \]

Step 2: Apply proportionality theorem. Since: \[ DE \parallel BC \] we use: \[ \frac{AD}{AB} = \frac{AE}{AC} \] Substitute known values: \[ \frac{2}{5} = \frac{AE}{2.5} \]

Step 3: Solve for \(AE\). Cross multiply: \[ 5 \times AE = 2 \times 2.5 \] \[ 5AE = 5 \] Divide both sides by 5: \[ AE = 1 \]

Step 4: Verification step. Since: \[ AC = 2.5 \] then: \[ EC = 2.5 - 1 = 1.5 \] Now check ratio: \[ \frac{AE}{EC} = \frac{1}{1.5} = \frac{2}{3} \] which matches: \[ \frac{AD}{DB} = \frac{2}{3} \] Hence answer is verified. Final Answer: \[ \boxed{1 \text{ cm}} \]