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' \[DE=DF\] |
\[\therefore \] \[AB=AC\] |
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