sir, please answer this question......
Asked by sahithi_16 | 15th Nov, 2009, 10:44: AM
Construction: Join EF
Since the line segment joining the mid points of two sides of a triangle is parallel to the third side.
We have, EF || BC
Triangles BEF and CEF are the same base EF and between the same parallel lines.
So, ar(BEF) = ar(CEF)
i.e ar(BEF) - ar(GEF) = ar(CEF) - ar (GEF)
ar (BFG) = (CEG)
We know that the median of a triangle divides it into two triangles of equal area.
So, ar(BEC) = ar(ABE)
i.e ar(BGC) + ar(CEG)= ar(quad.AFGE) + ar(BFG)
i.e ar(BGC) + ar(BFG)= ar(quad.AFGE) + ar(BFG)
Therefore, i.e ar(BGC) = ar(quad.AFGE)
Answered by | 11th Dec, 2009, 11:54: AM
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