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 |
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