the medians of be and cf of triangle abc intersects at G. prove area of gbc = area of quadrilateral afge
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triangle on same base and between same parallel lines are equal in area so
ar. BCF = ar. BCE
ar. BCG+ CEG =ar. BCG + BFG
ar. CEG = ar. BFG...........(1)
the median of triangle divide it into two triangle of equle area .
BE is median of triangle ABC
ar. BCE = ar. ABE
ar. BCG + CEG = ar. BFG + AFGE
ar. BCG + CEG = ar. CEG + AFGE [frm eq. 1]
ar..BCG = ar. AFGE
ar. BCF = ar. BCE
ar. BCG+ CEG =ar. BCG + BFG
ar. CEG = ar. BFG...........(1)
the median of triangle divide it into two triangle of equle area .
BE is median of triangle ABC
ar. BCE = ar. ABE
ar. BCG + CEG = ar. BFG + AFGE
ar. BCG + CEG = ar. CEG + AFGE [frm eq. 1]
ar..BCG = ar. AFGE
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Answer:
According to question the area of ΔGBC = area of the quadrilateral AFGE.
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