Question

Let AB and CD be two parallel lines and PQ be a transversal. Show that the angle bisector of a pair of two internal angles on the same side of the transversal are perpendicular to each other

Answer 1

Answer:

As shown in the figure, AB and CD are two parallel lines and PQ is a transverse line which meets AB on R and CD on S. ∠BRS and ∠RSD are two internal angles on the same side, the bisectors of the angles ∠BRS and ∠RSD respectively meet at point T

To prove that: RT⊥TS

Proof:

∵ RT is angle bisector of ∠BRS

∴ ∠SRT = ∠BRT = (1/2)∠BRS

Similarly,

∠RST = ∠DST = (1/2)∠RSD

∵ AB ║ CD and PQ is transverse line

∴ ∠BRS + ∠RSD = 180°   (sum of two internal angles of the same side of parallel lines)

Multiplying by 1/2 on both the sides

(1/2)∠BRS + (1/2)∠RSD = (1/2)180°

or, ∠SRT + ∠RST = 90°

Therefore, in Δ RTS

∠SRT + ∠RST + ∠RTS =  180°  (sum of internal angles of a triangle)

or, 90° + ∠RTS =  180°

⇒ ∠RTS =  90°

∴ RT⊥TS            (Proved)

Answer 2

Answer:

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