In a $△ABC$, $DE\parallel BC$ such that $D$ lies on $AB$ and $E$ lies on $AC$. If $AD=3.6\text{ cm}$, $DB=2.4\text{ cm}$, and $EC=3\text{ cm}$, find the length of $AE$.

Step-by-step solution

1. According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. For $△ABC$ with $DE\parallel BC$, we write the ratio of the divided segments as: $\frac{AD}{DB}=\frac{AE}{EC}$.
3. Substitute the given values $AD=3.6\text{ cm}$, $DB=2.4\text{ cm}$, and $EC=3\text{ cm}$ into the equation: $\frac{3.6}{2.4}=\frac{AE}{3}$.
4. Simplify the left-hand side fraction: $\frac{3.6}{2.4}=\frac{3}{2}$.
5. Solve for $AE$: $AE=\frac{3}{2}\times 3=4.5\text{ cm}$.

Concept tested: Basic Proportionality Theorem (Thales Theorem)

Published 20 August 2026 · How this solution was prepared and checked

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