Since (DE \parallel BC), triangles (ADE) and (ABC) are similar.
Let:
[
\text{Area}(ADE) : \text{Area}(DECB) = 4 : 21
]
So total area:
[
\text{Area}(ABC) = 4 + 21 = 25 \text{ parts}
]
Thus:
[
\frac{\text{Area}(ADE)}{\text{Area}(ABC)} = \frac{4}{25}
]
Step 1: Use similarity ratio
For similar triangles:
[
\frac{\text{Area}(ADE)}{\text{Area}(ABC)} = \left(\frac{AD}{AB}\right)^2
]
So:
[
\left(\frac{AD}{AB}\right)^2 = \frac{4}{25}
]
[
\frac{AD}{AB} = \frac{2}{5}
]
Step 2: Find AD : DB
If (AD/AB = 2/5), then:
(AD = 2x)
(AB = 5x)
(DB = AB - AD = 3x)
So:
[
AD : DB = 2x : 3x = 2 : 3
]
✅ Final Answer:
[
\boxed{2:3}
]