Răspuns:
Explicație pas cu pas:
ABCD trapez isoscel circumscriptibil, AB=B, CD=b.
a) x=AE=BH, unde BH⊥AD, y=CF, CF⊥AD. deci BH║CF, ⇒ ΔDCF~ΔABH, ⇒ CF/BH=DC/AB, ⇒ y/x=b/B.
b) Ttrasăm DN⊥AB, AM=AG=B/2, DM1=DG=b/2, unde M și M1 puncte de tangență. Atunci AD=(B+b)/2. DM1=MN=b/2, deci AN=(B-b)/2. Din ΔADN, ⇒
[tex]DN^2=AD^2-AN^2=(\dfrac{B+b}{2})^2- (\dfrac{B-b}{2})^2=(\dfrac{B+b}{2}+ \dfrac{B-b}{2})(\dfrac{B+b}{2}-\dfrac{B-b}{2})=Bb, ~~deci~DN=\sqrt{Bb}.\\[/tex]
ΔABH~ΔADN, dreptunghice cu unghi ascuțit comun. ⇒
[tex]\dfrac{AB}{AD}=\dfrac{BH}{DN} ~=>~\dfrac{B}{\frac{B+b}{2} }=\dfrac{x}{ \sqrt{Bb}},~=>~x=\dfrac{B\sqrt{Bb} }{\frac{B+b}{2} },~=>~x=\dfrac{2B\sqrt{Bb} }{B+b}\\ Din~\dfrac{y}{x}=\dfrac{b}{B},~=>~y=\dfrac{xb}{B}=\dfrac{\frac{2B\sqrt{Bb} }{B+b}b}{B}=\dfrac{2b\sqrt{Bb} }{B+b} \\[/tex]
