[tex]\it CF = 6=\dfrac{2}{3}\cdot9= \dfrac{2}{3}BC
\\\;\\
AE = 16=\dfrac{2}{3}\cdot24= \dfrac{2}{3}AB
\\\;\\
\mathcal{A}_{ABCD} = AB\cdot AD\cdot sinA = CD\cdot BC\cdot sinC
[/tex]
[tex]\it \mathcal{A}_{ADE} = \dfrac{1}{2}\cdot AE\cdot AD\cdot sinA=\dfrac{1}{2}\cdot \dfrac{2}{3}AB\cdot AD\cdot sinA = \dfrac{1}{3}\mathcal{A}_{ABCD}
\\\;\\ \\\;\\
\mathcal{A}_{DCF} = \dfrac{1}{2}\cdot CF\cdot CD\cdot sinA=\dfrac{1}{2}\cdot \dfrac{2}{3}BC\cdot CD\cdot sinA = \dfrac{1}{3}\mathcal{A}_{ABCD} [/tex]
Dacă ariile celor două triunghiuri reprezintă 1/3 din aria paralelogramului,
atunci aria patrulaterului DEBF reprezintă tot o treime din aria (ABCD).
Prin urmare, avem :
[tex]\it \mathcal{A}_{ADE } = \mathcal{A}_{DCF} = \mathcal{A}_{DEBF } = \dfrac{1}{3}\mathcal{A}_{ABCD}[/tex]
