Ai desenul atasat. Construim CF perpendicular pe AE.
[tex] \frac{CE}{ED} [/tex]=0,5 adica:
[tex] \frac{CE}{ED} [/tex]=[tex] \frac{5}{10} [/tex]
[tex] \frac{CE}{ED} [/tex]=[tex] \frac{1}{2} [/tex] si folosim proprietatile de la proportii:
[tex] \frac{CE}{CE+ED} [/tex]=[tex] \frac{1}{1+2} [/tex]
[tex] \frac{CE}{CD} [/tex]=[tex] \frac{1}{3} [/tex]
[tex] \frac{CE}{10} [/tex]=[tex] \frac{1}{3} [/tex] deci:
CE=[tex] \frac{10}{3} [/tex] cm, iar
ED=10 - [tex] \frac{10}{3} [/tex]
ED=[tex] \frac{20}{3} [/tex]
Observam ca in ΔCFE si ΔADE avem unghiurile <(CEF)≡<(AED) ca unghiuri opuse la varf, iar m(<ADE)=m(<CFE)=90 grade, deci
ΔCFE ≈ ΔADE (cazul U.U.) si avem rapoartele egale:
[tex] \frac{CF}{AD} = \frac{CE}{AE} [/tex], unde
AD=10 cm
CE=[tex] \frac{10}{3} [/tex] cm
AE il calculam cu Teorema lui Pitagora in triunghiul ADE dreptunghic:
[tex] AE^{2} = AD^{2} + ED^{2} [/tex]
[tex] AE^{2} = 10^{2} + ( \frac{20}{3} )^{2} [/tex]
[tex] AE^{2} = 100 + \frac{400}{9} [/tex]
[tex] AE^{2} = \frac{1300}{9} [/tex]
AE=[tex] \frac{10 \sqrt{13} }{3} [/tex] cm, deci revenim in relatia de mai sus:
CF=[tex] \frac{AD*CE}{AE} [/tex]
CF=[tex] \frac{10* \frac{10}{3} }{ \frac{10 \sqrt{13} }{3} } [/tex]
CF=[tex] \frac{10}{ \sqrt{13} } [/tex]
CF=[tex] \frac{10 \sqrt{13} }{13} [/tex] cm
