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Q.

Is the following statement true or false?


A ΔABC   its side BC   is bisected at D  . O   is any point on AD  . BO  and CO  are produced meet AC  and AB   at E   and F   respectively. AD   is produced to X   so that D   is the midpoint of O X  . Then AO:AX=AF:AB   and EFBC  .


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a

True

b

False 

answer is A.

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Detailed Solution

Given that, BC   is bisected at point D  .
Joining B-X and C-X .
Question Image BD=DC OD=OX   The diagonals OX  and BC  of quadrilateral BOCX  bisect each other. Therefore BOCX   is a parallelogram.
BOCX   and BXCO  
  BXCF   and CXBE   BXOF   and CXOE  
By using Thales' theorem in ΔABX  ,
AO AX = AF AB    …..i)
Now taking ΔACX   , where, CXOE  .
Again using Thales' theorem,
AO AX = AE AC  …..ii)
Solving equation i) and ii),
AO AX = AE AC  
By applying the converse of the theorem in ΔABC    we get EFCB   which is true.
Therefore the correct option is 1.
 
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