A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that

(i) ACBD is a rectangle.

(ii) CD is parallel to the original parallel lines.

From figure,

LM || PQ and AB is transversal.

AC, AD, BC and BD are bisectors of ∠LAB, ∠BAM, ∠PBA and ∠ABQ respectively.

So,

∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8.

(i) ∠LAB + ∠BAM = 180° (As LAM is a straight line)

$\dfrac{1}{2}$(∠LAB + ∠BAM) = $\dfrac{1}{2}$ x 180°

∠2 + ∠3 = 90° [Since, AC and AD are bisector of ∠LAB and ∠BAM]

∠CAD = 90°

∠A = 90°.

∠PBA + ∠QBA = 180° (As PBQ is a straight line)

$\dfrac{1}{2}$(∠PBA + ∠QBA) = $\dfrac{1}{2}$ x 180°

∠6 + ∠7 = 90° [Since, BC and BD are bisector of ∠PBA and ∠QBA]

∠CBD = 90°

∠B = 90°.

∠LAB + ∠ABP = 180° (LM || PQ, sum of co-interior angles is 180°)

$\dfrac{1}{2}$(∠LAB + ∠ABP) = $\dfrac{1}{2}$ x 180°

∠2 + ∠6 = 90° [Since, AC and BC are bisector of ∠LAB and ∠PBA]

In △ABC,

⇒ ∠2 + ∠6 + ∠C = 180°

⇒ 90° + ∠C = 180°

⇒ ∠C = 90°.

∠MAB + ∠ABQ = 180° (LM || PQ, sum of co-interior angles is 180°)

$\dfrac{1}{2}$(∠MAB + ∠ABQ) = $\dfrac{1}{2}$ x 180°

∠3 + ∠7 = 90° [Since, AD and BD are bisector of ∠MAB and ∠ABQ]

In △ABD,

⇒ ∠3 + ∠7 + ∠D = 180°

⇒ 90° + ∠D = 180°

⇒ ∠D = 90°.

From figure,

∠BAM = ∠ABP (Alternate angles are equal)

$\dfrac{∠\text{BAM}}{2} = \dfrac{∠\text{ABP}}{2}$

∠3 = ∠6

∠LAB = ∠ABQ (Alternate angles are equal)

$\dfrac{∠\text{LAB}}{2} = \dfrac{∠\text{ABQ}}{2}$

∠2 = ∠7

In △ABC and △ABD,

∠3 = ∠6 (Proved above)

∠2 = ∠7 (Proved above)

AB = AB (Common)

Hence, △ABC ≅ △ABD by ASA axiom.

∴ AD = BC and AC = BD (By C.P.C.T.)

Since, ∠A = ∠B = ∠C = ∠D = 90° and AD = BC, AC = BD.

Hence, proved that ACBD is a rectangle.

(ii) In △OAD,

OA = OD (Diagonals of rectangle bisect each other)

∠3 = ∠9 (Angles opposite to equal side are equal)

Since, ∠3 = 4,

∠9 = ∠4

Since, ∠9 and ∠4 are alternate angles and since they are equal we can say that,

⇒ OD || LM

⇒ CD || LM

Since, LM || PQ and CD || LM

⇒ CD || PQ.

Hence, proved that CD is parallel to the original parallel lines.