Question

In a parallelogram LEFT, the bisector of angle L also bisects EF in P. Prove that LT = 2LE.

Answer 1

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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