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In triangle ABC DE parallel to BC, BE and CD intersect at F, AD:BD = 5:4, find Ar(def)/ar(cfb)

Answer» Let AD = 5x cm and DB = 4x cm. Then AB = AD + DB = (5x + 4x) cm = 9x cm.In\xa0{tex}\\triangle ADE{/tex}\xa0and\xa0{tex}\\triangle ABC{/tex}, we have{tex}\\angle A D E = \\angle A B C{/tex}\xa0[corresponding angles]{tex}\\angle A E D = \\angle A C B{/tex}\xa0[corresponding angles]{tex}\\angle A = \\angle A {/tex}\xa0[common in both triangles]\xa0{tex}\\therefore \\quad \\triangle A D E \\sim \\triangle A B C{/tex}\xa0[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]\xa0{tex}= \\frac { 5 x } { 9 x } = \\frac { 5 } { 9 }{/tex}In\xa0{tex}\\triangle DFE{/tex}\xa0and\xa0{tex}\\triangle CFB{/tex}\xa0, we have{tex}\\angle E D F = \\angle B C F{/tex}\xa0[alternate interior angles as DE|| BC ]and\xa0{tex}\\angle D E F = \\angle C B F{/tex}\xa0[alternate interior angles as DE || BC].\xa0{tex}\\therefore \\quad \\triangle D F E \\sim \\triangle C F B{/tex}\xa0( By AA similarity Criteria)\xa0{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.


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