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.

Question

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.

KnowledgeBoat · Accepted Answer

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) (∠LAB + ∠BAM) = 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) (∠PBA + ∠QBA) = 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°) (∠LAB + ∠ABP) = 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°) (∠MAB + ∠ABQ) = 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) ∠3 = ∠6 ∠LAB = ∠ABQ (Alternate angles are equal) ∠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.

Mathematics

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.

Rectilinear Figures

Related Questions

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that
(i) BE bisects ∠B
(ii) ∠AEB = a right angle.

Mathematics

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.

Rectilinear Figures

Related Questions

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that(i) BE bisects ∠B(ii) ∠AEB = a right angle.

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.

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that
(i) BE bisects ∠B
(ii) ∠AEB = a right angle.