SSC CGL Tier II 12 Apr 2015 Paper

Question : 29 of 100

Marks: +1, -0

ABC is a triangle in which DE || BC and AD : DB = 5 : 4. Then DE : BC is

Solution:

(4)

DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA – similarly,
ΔADE ~ ΔABC

∴{A B}/{A D}= {B C}/{D E}

∴{D B}/{A D}=\ {4}/{5}

⇒{D B}/{A D}+1= {4}/{5}+1

⇒{D B+A D}/{A D}= {4+5}/{5}

⇒{A B}/{A D}= {9}/{5}= {B C}/{D E}

∴{D E}/{B C}= {5}/{9}

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