In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP || FQ .
It is given that, AB || CD and t is a transversal.
∴ ∠AEF = ∠EFD .....(1) (Pair of alternate interior angles)
EP is the bisectors of ∠AEF. (Given)
$\therefore \angle \mathrm{AEP}=\angle \mathrm{FEP}=\frac{1}{2} \angle \mathrm{AEF}$
⇒ ∠AEF = 2∠FEP .....(2)
Also, FQ is the bisectors of ∠EFD.
$\therefore \angle \mathrm{EFQ}=\angle \mathrm{QFD}=\frac{1}{2} \angle \mathrm{EFD}$
⇒ ∠EFD = 2∠EFQ .....(3)
From (1), (2) and (3), we have
2∠FEP = 2∠EFQ
⇒ ∠FEP = ∠EFQ
Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal.
∴ EP || FQ (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
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