A) \[AD=AB+BC+AC\]
B) \[2AD=AB+BC+AC\]
C) \[AD=\frac{1}{2}(AB+BC+AC)\]
D) \[3AD=AB+BC+AC\]
Correct Answer: B
Solution :
As the tangents drawn to a circle from a point outside It are equal, we have\[AD=AE\], \[BD=BF\]and\[CE=CF\] \[AD=AB+BD=AB+BF\] \[AD=AC+CE=AC+CF\] \[\therefore \]\[2AD=AB+AC+BF+CF\] \[2AD=AB+AC+BC\]You need to login to perform this action.
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