Saudi Arabia Mathematical Competitions

a. $\widehat{FEC}=180^{\circ }-\widehat{AEF}-\widehat{DEA}$ $$=180^{\circ }-\widehat{ADE}-\widehat{DEA}=\widehat{DAE}$$ From $AD=DC=CB<AB$ and $AB\parallel CD$ it follows that $\widehat{ADC}=\widehat{DCB}$, hence triangles $ADE$ and $ECF$ are similar. This yields $$\frac{AD}{EC}=\frac{AE}{EF}=\frac{DE}{CF}.\text{\hspace{1em}(1)}$$ This leads to $$AD\cdot CF=EC\cdot DE\le \frac{1}{4}(EC+DE{)}^2=\frac{1}{4}C{D}^2,$$ hence $4CF\le CB$, because $AD=DC=CB$.



b. If $4CF=CB$, then the inequality $$EC\cdot DE\le \frac{1}{4}(EC+DE{)}^2$$ becomes an equality, that is $CE=ED$.

Relation (1) becomes $\frac{AD}{DE}=\frac{AE}{EF}$. Since $\widehat{ADE}=\widehat{AEF}$, triangles $ADE$ and $AEF$ are similar.

Then $\widehat{DAE}=\widehat{EAF}$, so $AE$ bisects the angle $\widehat{DAF}$.