[tex]\it Teorema\ bisectoarei \Rightarrow \dfrac{BD}{CD}=\dfrac{AB}{AC} \Rightarrow \dfrac{BD}{CD}=\dfrac{\ 21^{(3}}{15} \Rightarrow\dfrac{BD}{CD}=\dfrac{7}{5} \Rightarrow\\ \\ \\ \Rightarrow\dfrac{BD}{CD}=1,4 \Rightarrow BD=1,4CD\ \ \ \ \ (1) \\ \\ \\ Dar,\ BD+CD=BC \Rightarrow BD+CD=24\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow 1,4CD+CD+24 \Rightarrow 2,4CD=24 \Rightarrow CD=10\ cm[/tex]
[tex]\it BD=BC-CD=24-10=14\ cm\\ \\ \\ b)\ DE||AB \stackrel{T.f.a.}{\Longrightarrow}\ \Delta ABC\sim\Delta EDC \Rightarrow \dfrac{\mathcal{P}_{ABC}}{\mathcal{P}_{EDC}}=k\ (raportul\ de\ asem\breve{a}nare )\ \ \ (*)\\ \\ \\ \Delta ABC\sim\Delta EDC \Rightarrow k=\dfrac{BC}{CD}=\dfrac{24}{10}=2,4[/tex]
Acum, relația (*) devine:
[tex]\it \dfrac{21+24+15}{\mathcal{P}_{EDC}}=2,4 \Rightarrow \dfrac{60}{\mathcal{P}_{EDC}}=2,4 \Rightarrow \mathcal{P}_{EDC}=\dfrac{^{5)}60}{2,4}=\dfrac{300}{12}=25\ cm[/tex]
