In interiorul pâtratului ABCD se aflà punctul E astfel incât triunghiul EAB este echilateral, de laturá AB=6сm
(2p) a) Sã se arate cã m(CDE) = 15°
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∆ABE echilateral => m(EAB) = 60°
m(DAE) = m(DAB) - m(EAB) = 90° - 60° = 30°
AE = AD = 6 cm => ∆AED isoscel => m(AED) = m(ADE) = x°
m(DAE) + m(AED) + m(ADE) = 180° =>
=> x° + x° = 180° - 30° =>
=> x° = 75° => m(ADE) = 75°
m(CDE) = m (ADC) - m(ADE) = 90° - 75° = 15°