In ΔABC, a line parallel to BC divides AB and AC such that the smaller triangle cut off at vertex A has area $\frac{1}{4}$ of ΔABC. What is the ratio in which the line divides AB (from A)?

**Theorem**: The ratio of the areas of two similar triangles is equal to the SQUARE of the ratio of their corresponding sides.

If ΔABC ~ ΔDEF with scale factor $k=AB/DE$, then
Area(ABC)/Area(DEF) = $k^2=(AB/DE{)}^2$.

**Why**: area scales as length × length. If both length and breadth scale by $k$, area scales by $k^2$.

**Example**: Two similar triangles have corresponding sides in ratio 3:5. Area ratio = 9:25.

**Reverse**: if areas are in ratio 16:25, sides are in ratio 4:5 (square root).