Q7 of 45 Page 164

Let and be two parallel lines and be a transversal. Let intersect in L. Suppose the bisector of ∠ALP intersect in R and the bisector of ∠PLB intersect in S. Prove that ∠LRS + ∠RSL = 90°.

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We know, ∠ALP + ∠BLP = 180°


According to problem,


∠ALR = ∠RLP = ∠ALD/2


Similarly,


∠BLS = ∠SLP = ∠BLP/2


∴ ∠RLP + ∠SLP = ∠ALD/2 + ∠BLD/2


⇒ ∠RLS = (∠ALP + ∠BLP)/2 = 180° = 90°


∴ ∠LRS + ∠RSL = 180° - ∠RLS


∠LRS + ∠RSL =180° - 90°


=90°


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