Quantitative Aptitude Ques 661
Question: In the given figure, D is the mid-point of $ BC, $ $ DE\bot AB $ and $ DF\bot AC $ such that DE = DF. Then, which of the following is true?
Options:
A) AB = AC
B) AC = BC
C) AB = BC
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \Delta DBE\sim \Delta DCF $
$ \frac{DB}{DC}=\frac{BE}{CF}=\frac{DE}{DF} $
But $ DE=DF $
$ \Rightarrow $ $ BE=CF $ In $ \Delta BDE\sim \Delta BCA, $
$ \Rightarrow $ $ \frac{BD}{BC}=\frac{DE}{CA} $
$ \Rightarrow $ $ \frac{BD}{2BD}=\frac{DE}{AC} $
$ \Rightarrow $ $ AC=2DE $
Similarly, using $ \Delta CFD\sim \Delta CAB, $ $ AB=2DF $
But’ $ DE=DF $
$ \therefore $ $ AB=AC $
BETA
