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In chapter similar triangles,i have doubt in the second question of smartindia class note. How we will get N IN THE FIGURE


  Mid points of sides BC;  AC and AB of   ?ABC are respectively D,E and F. So lines AD, BE and CF are medians of the triangle.
G is the point of intersection of these medians.  So G is the centroid of  ? ABC.  The centroid divides each median of a triangle in the ratio 2 : 1.
  ? AG : GD = 2 : 1
Perpendicular from  A to opposite side BC  cuts BC at M  and the parallel line to BC through G at N.
So AN : NM = 2 :1
 ?AM = AN + NM = 2 NM + NM = 3NM.
Area of   ?ABC = 1/2 BC  x AM.
                           = 1/2 BC x 3NM  = 3 x 1/2  BC x NM
Area of    ?ABC = 1/2 BC x NM
? Area of   ?ABC = 3 x area of  ?BGC.
ie  Area of  ?BGC = 1/3 area of   ?ABC.
 Like  this we can show that areas of   ?AGC and   ?AGB
are equal to 1/3 are of   ?ABC.
So areas of   ?AGB,  ?AGC and   ?BGC are equal.


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