Question

If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
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Answer 1

$\sf \Large Solution!!$

$\sf \large To\:prove:$

$\sf \to PQ\parallel RS$

$\sf \large Now,$

$\sf \to \angle ABE=\frac{1}{2}\angle ABQ$

$\sf \to \angle BCG=\frac{1}{2}\angle BCS$

$\sf \to BE\parallel CG$

$\sf \therefore \angle ABE=\angle BCG\quad (Corresponding\:angles)$

$\sf \to \cancel{\frac{1}{2}}\angle ABQ=\cancel{\frac{1}{2}}\angle BCS$

$\sf \to \angle ABQ=\angle BCS$

$\sf \boxed{\therefore PQ\parallel RS}$

Thanks for asking!! :)

Answer 2

Step-by-step explanation:

Solution!!

\sf \large To\:prove:Toprove:

\sf \to PQ\parallel RS→PQ∥RS

\sf \large Now,Now,

\sf \to \angle ABE=\frac{1}{2}\angle ABQ→∠ABE=

2

1

∠ABQ

\sf \to \angle BCG=\frac{1}{2}\angle BCS→∠BCG=

2

1

∠BCS

\sf \to BE\parallel CG→BE∥CG

\sf \therefore \angle ABE=\angle BCG\quad (Corresponding\:angles)∴∠ABE=∠BCG(Correspondingangles)

\sf \to \cancel{\frac{1}{2}}\angle ABQ=\cancel{\frac{1}{2}}\angle BCS→

2

1

∠ABQ=

2

1

∠BCS

\sf \to \angle ABQ=\angle BCS→∠ABQ=∠BCS

\sf \boxed{\therefore PQ\parallel RS}

∴PQ∥RS