THE MEDIANS BE & CF of a triangle ABC intersect at G . prove that area of triangle GBC = area of quadrilateral AFGE .figure is must . PLZ help me !!!!!
hrishisatya:
and also AB = AC ( GIVEN )
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given ,
BE & CF are medians
∴ E and F are midpoints of AC & AB
∴ ΔBCE = ΔBEA ----------- ( i )
ΔBCF = ΔCAF
Const. join E to F we get FE // BC ( By midpoint theoram i.e E joind to F)
∴ ΔFBC = ΔBCE ( Δ on the same base between same // are equal in area )
ΔFBC - ΔGBC = ΔBCE - ΔGBC
⇒ ΔCGE = ΔFGE ( GBC is common ) ----------------- ( ii )
on subtracting eqn. ( i ) & ( ii ) we get ,
ΔBCE - ΔCGE = ΔBEA - ΔFGE
∴ ΔBGC =quadrilateral AFGE .
BE & CF are medians
∴ E and F are midpoints of AC & AB
∴ ΔBCE = ΔBEA ----------- ( i )
ΔBCF = ΔCAF
Const. join E to F we get FE // BC ( By midpoint theoram i.e E joind to F)
∴ ΔFBC = ΔBCE ( Δ on the same base between same // are equal in area )
ΔFBC - ΔGBC = ΔBCE - ΔGBC
⇒ ΔCGE = ΔFGE ( GBC is common ) ----------------- ( ii )
on subtracting eqn. ( i ) & ( ii ) we get ,
ΔBCE - ΔCGE = ΔBEA - ΔFGE
∴ ΔBGC =quadrilateral AFGE .
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