Exercise 1 Page 451

Let's begin by recalling the diagram given in the mentioned exercise. Here, DEFG is a rhombus, so all its sides are congruent.

Notice that △ DEF is an isosceles triangle. Then, by the Base Angles Theorem, we have that ∠ EDF ≅ ∠ DFE, which implies that m∠ DFE = 48^(∘).

Next, by the Side-Side-Side (SSS) Congruence Theorem, we get that △ DEF ≅ △ DGF. This implies that ∠ EDF ≅ ∠ 4 and ∠ DFE≅ ∠ 5, so m ∠ 4 = 48^(∘) and m ∠ 5 = 48^(∘).

That way, we've found the required angles without using the Rhombus Opposite Angles Theorem.