State nd proof bpt and hence show that the diagonals of a trapazium each other propotionaly
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So, diagonals of a trapezium divide each other proportionally.
DE/EB = CE/EA
From (1) and (2), we get
CE/EA = DF/FA [According to basic proportionality theorem] --------- (2)
FE || DC (Given)
DE/EB = DF/FA [According to basic proportionality theorem] --------- (1)
FE || AB (Given)
Draw EF || BA || CD, intersecting AD in F.
Proof:
DE/EB = CE /EA
To Prove:
ABCD is a trapezium.
Given:
DE/EB = CE/EA
From (1) and (2), we get
CE/EA = DF/FA [According to basic proportionality theorem] --------- (2)
FE || DC (Given)
DE/EB = DF/FA [According to basic proportionality theorem] --------- (1)
FE || AB (Given)
Draw EF || BA || CD, intersecting AD in F.
Proof:
DE/EB = CE /EA
To Prove:
ABCD is a trapezium.
Given:
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