In the figure, D,E and F are respectively the midpoints of sides AB, BC and AC of triangle ABC. CG is drawn parallal to BA and it meets DF produced in G. Prove that DEF congruent to FCG.
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DF is parallel to BC(mid point theorem).
hence DF//EC.
DF IS PRODUCED TO G.
THEREFORE FG//EC (1)
FE//AB(MID POINT THEOREM)
CG//AB(GIVEN)
THEREFORE FE//CG(2)
SO, EFGC IS A PARALLELOGRAM( FROM 1 AND 2)
IN ΔDEF AND ΔFCG,
FE=GC ( OPPOSITE SIDES OF A PARALLELOGRAM)
FC=1/2 AC( F IS MID POINT OF AC)
DE=1/2AC(MID POINT THEOREM)
SO FC=DE
FG=EC( OPPOSITE SIDES OF A PARALLELOGRAM)
DF=1/2BC=EC (MID POINT THEOREM AND E IS MID POINT OF BC)
SO FG= DF
HENCE ΔDEF ~=ΔFCG(S.S.S. TEST)
hence DF//EC.
DF IS PRODUCED TO G.
THEREFORE FG//EC (1)
FE//AB(MID POINT THEOREM)
CG//AB(GIVEN)
THEREFORE FE//CG(2)
SO, EFGC IS A PARALLELOGRAM( FROM 1 AND 2)
IN ΔDEF AND ΔFCG,
FE=GC ( OPPOSITE SIDES OF A PARALLELOGRAM)
FC=1/2 AC( F IS MID POINT OF AC)
DE=1/2AC(MID POINT THEOREM)
SO FC=DE
FG=EC( OPPOSITE SIDES OF A PARALLELOGRAM)
DF=1/2BC=EC (MID POINT THEOREM AND E IS MID POINT OF BC)
SO FG= DF
HENCE ΔDEF ~=ΔFCG(S.S.S. TEST)
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