SATHEE: Quantitative Aptitude Ques 661

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

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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:

Answer:

Solution:

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