Cu formula lui Heron, calculam aria triunghiului ABC.
Formula este [tex]S=\sqrt{p(p-a)(p-b)(p-c)},\ unde \ p=\dfrac{a+b+c}{2}[/tex]
Se obtine [tex]p=9\ cm;\ S=6\sqrt6\ cm^2[/tex]
Din teorema bisectoarei avem: [tex]\dfrac{AD}{DB}=\dfrac{AC}{BC}[/tex], pe care o vom folosi mai departe:
[tex]\dfrac{A_{ADC}}{A_{BDC}}=\dfrac{AD\cdot d(C.AB)}{DB\cdot d(C,AB)}=\dfrac{AC}{BC}=\dfrac57[/tex]. Adunam numaratorii la numitori, si obtinem:
[tex]\dfrac{A_{ADC}}{A_{ABC}}=\dfrac{5}{12}\Rightarrow A_{ADC}=S\cdot\dfrac{5}{12}=\dfrac{5\cdot6\sqrt6}{12}=\dfrac{5\sqrt6}{2}\ cm^2[/tex]
