Answer:
[tex]\frac{4}{10}=\frac{3.2}{8.0}[/tex]
Step-by-step explanation:
Given,
ABC is a triangle,
In which, AD = 4 unit, DB = 6 unit, AE = 3.2 unit and EC = 4.8 unit,
⇒ [tex]\frac{AD}{AB}=\frac{4}{4+6}=\frac{4}{10}=0.4[/tex]
[tex]\frac{AE}{EC}=\frac{3.2}{3.2+4.8}=\frac{3.2}{8}=0.4[/tex]
[tex]\implies \frac{AD}{AB}=\frac{AE}{EC}[/tex]
⇒ Δ ADE ≅ Δ ABC
⇒ ∠ADE ≅ ∠ABC and ∠AED ≅ ∠ACB
By the converse of alternate interior angle theorem.
BC ║ DE
Hence, the required proportion of segment is,
[tex]\frac{4}{10}=\frac{3.2}{8.0}[/tex]
