In a parallelogram LEFT, the bisector of angle L also bisects EF in P. Prove that LT = 2LE.
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Given : In a parallelogram LEFT, the bisector of angle L also bisects EF in P.
To Find : Prove that LT = 2LE.
Solution:
LEFT is parallelogram
=> LT || EF and LP is transversal
=> ∠EPL = ∠TLP ( alternate angle)
∠ELP = ∠TLP ( as LP is bisector of ∠L
=> ∠EPL = ∠ ELP
in Δ EPL
∠EPL = ∠ ELP
Hence EP = LE
EP = (1/2)EF
=> (1/2)EF = LE
=> EF = 2LE
EF = LT ( as opposite side of parallelogram are equal)
=> LT = 2LE
QED
Hence Proved
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