question_answer

In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively If AB = 2AD, then DE : BC is [SSC CGL Tier II, 2014]

A) 2 : 3

B) 2 : 1

C) 1 : 2

D) 1 : 3

Correct Answer: C

Solution :

[c] Now, AB = 2AD or \[AD=\frac{AB}{2}\] So, D is mid-point of AB Now, in \[\Delta \,ADE\]and \[\Delta \,ABC\] \[\angle \,A=\angle \,A\]              (common) \[\angle \,ADE=\angle ABC\,(DE\parallel BC)\] \[\therefore \]      \[\Delta \,ADE\simeq \Delta \,ABC\] \[\frac{AD}{AD}=\frac{DE}{BC},\]\[\frac{AB}{2\times AB}=\frac{DE}{BC},\]\[\frac{DE}{BC}=\frac{1}{2}\] DE : BC = 1 : 2