U konveksnom četverokutu ABCDABCDABCD vrijedi ∣AD∣=∣CD∣|AD| = |CD|∣AD∣=∣CD∣ i ∡ADC=90°\measuredangle ADC = 90°∡ADC=90°. Ako je ∣AB∣=a|AB| = a∣AB∣=a, ∣BC∣=b|BC| = b∣BC∣=b, ∣BD∣=d|BD| = d∣BD∣=d, ∡ABC=β\measuredangle ABC = \beta∡ABC=β, dokaži da vrijedi 2d2=a2+b2+2absinβ.2d^2 = a^2 + b^2 + 2ab \sin \beta.2d2=a2+b2+2absinβ.