Question

In `square`ABCD, side BC || side AD, side AB ≅ side DC If ∠A = 72° then find the measure of ∠B and ∠D.

Answer 1

Draw Seg BM ⊥ Seg AD such that A-M-D and Seg CN ⊥ Seg 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.)

In ∆BMA and ∆CND,

∠BMA = ∠CND = 90°

Hypotenuse BA ≅ Hypotenuse CD      ...(Given)

Seg BM ≅ Seg CN

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

∴ ∠BAM ≅ ∠CDN       ...(c.a.c.t)

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

∠BAD = 72°      ...(Given)

∠CDA = 72° i.e., ∠D = 72°

Seg BC || Seg AD       ...(Given)

 side BC || side AD and side BA is their transversal.

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

∴ 72° + ∠ABC = 180°

∴ ∠ABC = 180° – 72°

∴ ∠ABC = 108° i.e. ∠B = 108°