THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) The hypothesis gives $AC=\frac{1}{2}BC$ and $CD=\frac{1}{2}AC$, hence $CD=\frac{1}{4}BC$. From $△BDE\sim △DFE$ follows $\frac{BD}{DF}=\frac{DE}{FE}$, that is $\frac{DE}{BD}=\frac{EF}{FD}$. Now $\frac{DC}{DB}=\frac{1}{3}=\frac{AE}{DE}$ yields $\frac{DE}{BD}=\frac{AE}{DC}$, so $\frac{AE}{CD}=\frac{EF}{DF}$. Now $\widehat{AEF}=180^{\circ }-\widehat{DEF}=180^{\circ }-(90^{\circ }-\widehat{ADF})=90^{\circ }+\widehat{ADF}=\widehat{FDC}$. Therefore $△AEF\sim △CDF$ (SAS), whence $\widehat{AFE}\equiv \widehat{CFD}$. This leads to $\widehat{AFC}=$ $\widehat{AFE}+\widehat{EFC}=\widehat{CFD}+\widehat{EFC}=\widehat{EFD}=90^{\circ }$, hence $AF\perp FC$.



b) Since $\widehat{AFC}=90^{\circ }=\widehat{ADC}$, the quadrilateral $AFDC$ is cyclic, therefore $\widehat{DFC}=\widehat{DAC}=30^{\circ }$. Then $\widehat{AFB}=360^{\circ }-\widehat{BFD}-\widehat{DFC}-\widehat{CFA}=150^{\circ }$.