In triangle ABC DE parallel to …

Let AD = 5x cm and DB = 4x cm. Then AB = AD + DB = (5x + 4x) cm = 9x cm.
In {tex}\triangle ADE{/tex} and {tex}\triangle ABC{/tex}, we have
{tex}\angle A D E = \angle A B C{/tex} [corresponding angles]
{tex}\angle A E D = \angle A C B{/tex} [corresponding angles]

{tex}\angle A = \angle A {/tex} [common in both triangles] 
{tex}\therefore \quad \triangle A D E \sim \triangle A B C{/tex} [By AAA - Similarity Criteria ]
{tex}\Rightarrow \frac { D E } { B C } = \frac { A D } { A B }{/tex}[ As corresponding sides of similar triangles are proportionate to each other] 
{tex}= \frac { 5 x } { 9 x } = \frac { 5 } { 9 }{/tex}
In {tex}\triangle DFE{/tex} and {tex}\triangle CFB{/tex} , we have
{tex}\angle E D F = \angle B C F{/tex} [alternate interior angles as DE|| BC ]
and {tex}\angle D E F = \angle C B F{/tex} [alternate interior angles as DE || BC].
 {tex}\therefore \quad \triangle D F E \sim \triangle C F B{/tex} ( By AA similarity Criteria) 
{tex}\Rightarrow \quad \frac { \operatorname { ar } ( \Delta D F E ) } { \operatorname { ar } ( \Delta C F B ) } = \frac { D E ^ { 2 } } { C B ^ { 2 } } = \frac { D E ^ { 2 } } { B C ^ { 2 } }{/tex}{tex}= \left( \frac { D E } { B C } \right) ^ { 2 }{/tex}
{tex}= \left( \frac { 5 } { 9 } \right) ^ { 2 } = \frac { 25 } { 81 }{/tex}
{tex}\Rightarrow \quad \operatorname { ar } ( \triangle D F E ) : \operatorname { ar } ( \triangle C F B ){/tex}
= 25 : 81.