question_answer

E is the mid-point of the median AD of a triangle ABC. If BE produced meets the side AC at F, then CF is equal to?

A)  \[\frac{AC}{3}\]                          

B)  \[\frac{2AC}{3}\]

C)  \[\frac{AC}{2}\]                                              

D)  None of the above

Correct Answer: B

Solution :

 Draw line segment DG parallel to BF. Then, In  \[\Delta \,ADG,\text{ }EF||DG\] and \[AE=ED\text{  }\,\,\therefore \text{ }AF=GF\]    .....(i) Similarly, in \[\Delta BCF,\text{ }DG||BF\] and \[BD=DC\text{    }\therefore \text{ }FG=GC\]   .....(ii) From Eqs. (i) and (ii), CF \[\]