Question

$\mathcal \red{h} \orange{o} \pink{l} \purple{a}$



If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.

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Answer 1

$\underline \mathbb{SOLUTION}$

$\bold{\huge{\fbox{\color{Red}{Given}}}}$

》Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM , HM , GL and HL are the bisectors of the two pairs of interior angles.

$\bold{\huge{\fbox{\color{Red}{To\:Prove}}}}$

》GMHL is a rectangle

$\bold{\huge{\fbox{\color{Red}{Proof}}}}$

$\bold\green{\underline{Since,}}$

》AB || CD

》∠AGH = ∠DHG (alternate interior angles)

$\implies\:$ 1/2∠AGH = 1/2∠DGH

$\implies\:$ ∠1 = ∠2 (GM and HL are bisectors of angle AGH and angle DGH respectively).

$\implies\:$ GM || HL (∠1 and ∠2 form a pair of alternate interior angles and are equal)

$\bold\green{\underline{Similarly,}}$

$\implies\:$ GL || MH

$\bold {GHML\:is\:a\:parallelogram}$

$\bold\green{\underline{Since,}}$

$\implies\:$ AB || CD

$\implies\:$ ∠BGH + ∠DGH = 180° (sum of interior angles on the same side of transversal is 180 degree)

$\implies\:$ 1/2∠BGH + 1/2∠DHG = 90°

$\implies\:$ ∠3 + ∠2 = 90°

(GL and HL are bisectors of Angle BGH and angle DHG respectively).

$\boxed {In\:ΔGLH,}$

$\implies\:$ ∠2 + ∠3 + ∠L = 180°

$\implies\:$ 90° + ∠L = 180°

∠L = 180° - 90°

∠L = 90°

$\bold\green{\underline{Thus,}}$

》In parallelogram GMHL angle is 90°.

Hence, It is a rectangle

$\bold{\huge{\fbox{\color{Red}{Hence\:Proved}}}}$

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NOTE:

REFER TO ATTACHMENT

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Answer 2

$\huge{\underline{\mathbb{\red{Answer}}}}$

Refer to attachment for solution

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$\huge{\underline{\mathbb{\pink{Thankyou}}}}$