Question

In `square`ABCD, side BC < side AD in following figure. side BC || side AD and if side BA ≅ side CD then prove that ∠ABC ≅ ∠DCB.

Answer 1

Draw Seg BM ⊥ Seg AD such that A-M-D and Seg CN ⊥ AD so that A-N-D

Seg BC || Seg AD       ...(Given)

∴ BM = CN       ...(The distance between the lengths of the parallel sides of a parallelogram is equal.) ...(1)

In ∆BMA and ∆CND,

∠BMA = ∠CND = 90°

Hypotenuse BA ≅ Hypotenuse CD   ...(Given)

Seg BM ≅ Seg CN     ...[From (1)]

∴ ∆BMA ≅ ∆CND      ...(Hypotenuse side test)

∴ ∠BAM ≅ ∠CDN       ...(c.a.c.t)

That is, ∠BAD ≅ ∠CDA        ...(A-M-D, A-N-D) ...(2)

Seg BC || Seg AD         ...(Given)

and side AB is their transversal.

∴ ∠ABC + ∠BAD = 180°      ...(interior angle) ...(3)

Seg BC || Seg AD    ...(Given)

and side CD is their transversal.

∴ ∠DCB + ∠CDA = 180°      ...(interior angle) ...(4)

∴ ∠ABC + ∠BAD = ∠DCB + ∠CDA   ...[From (3) and (4)]

∴ ∠ABC + ∠CDA = ∠DCB + ∠CDA   ...[From (2)]

∴ ∠ABC = ∠DCB

∴ ∠ABC ≅ ∠DCB