question_answer

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?

A) AB = AC         

B) AC = BC

C) AB = BC                     

D) None of these

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'       \[DE=DF\]

\[\therefore \]      \[AB=AC\]