Find the corners of a rhombus in which one of the diagonals is equal to its side.

Let ABCD be a rhombus. All sides of the rhombus are equal, then AB = BC = CD = AD. AC and BD are the diagonals of the rhombus. Let the diagonal BD be equal to the side of the rhombus, then AB = BC = CD = AD = BD.
The diagonal BD divides the rhombus into two equal triangles: △ DAB = △ DCB.
Consider △ DAB: AD = AB = BD. Since all sides of △ DAB are equal, then this triangle is equilateral (regular), and all angles of a regular triangle are 60 °, then:
∠BDA = ∠DAB (aka ∠A) = ∠ABD = 60 °.
Since △ DAB = △ DCB, △ DCB is also an Ozette Triangle:
∠DBC = ∠BCD (aka ∠C) = ∠CDB = 60 °.
∠B consists of two corners:
∠B = ∠ABD + ∠DBC;
∠B = 60 ° + 60 ° = 120 °.
In a rhombus, the opposite angles are equal, then ∠B = ∠D = 120 °.
Answer: ∠A = 60 °, ∠B = 120 °, ∠C = 60 °, ∠D = 120 °.