In a quadrilateral ABCD, AB = CD and ∠B = ∠C. Prove that:

(i) AC = DB,

(ii) AD is parallel to BC.

(i) Given: In a quadrilateral ABCD, AB = CD and ∠B = ∠C.

To prove: AC = DB

Construction: Join diagonals AC and BD.

Proof: Consider the triangles ABC and DBC,

AB = CD (Given)

∠ABC = ∠BCD (Given)

BC = BC (Common Side)

Using SAS congruency criterion,

Δ ABC ≅ Δ DBC

By corresponding parts of congruent triangles,

Hence, AC = BD.

(ii) To prove: AD is parallel to BC.

Proof: Consider the triangles ABD and ADC,

AB = CD (Given)

AC = DB (Proved above)

DA = DA (Common Side)

Using SSS congruency criterion,

Δ ABD ≅ Δ ADC

By corresponding parts of congruent triangles,

∠A = ∠D

As we know that sum of all angles of quadrilateral is 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠D + ∠B + ∠B + ∠D = 360° [∠A = ∠D and ∠B = ∠C]

⇒ 2∠D + 2∠B = 360°

⇒ ∠D + ∠B = $\dfrac{360°}{2}$

⇒ ∠D + ∠B = 180°

Since, opposite angles formed between the transversal AC are supplementary.

Hence, AD is parallel to BC.