Since the three angles of ΔACD have equal measure,
ΔACD is equilateral, and ∠CAD = ∠CDA = ∠C = 60°
Therefore AC = 6 = AD = CD = y.
So y = 6
Since ∠C = ∠CAD = ∠CDA = 60°, ∠CAB = ∠CAD + ∠DAB = 60° + 30° = 90°
So ΔABC is a right triangle, and ∠B and ∠C are complementary,
So ∠B = 90°-∠C = 90°-60° = 30° = z°.
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We could have gotten z° from the fact that ΔABD is isosceles,
since AD = 6 and BD = 6, and the base angles are equal, so
∠B = ∠DAB = 30° = z².
Edwin
