It's a challenge, can you do it-MCQ

This result is known as Blanchet's Theorem.

Draw a line through $A$ parallel to $BC$ . Then the triangles $AME$ and $CDE$ are similar, as are the triangles $ANF$ and $BDF$ . Thus we deduce that $\frac{AE}{EC} \; = \; \frac{AM}{CD} \hspace{2cm} \frac{AF}{FB} \; = \; \frac{AN}{BD}$ Ceva's Theorem for the triangle $ABC$ tells us that $\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{AE} \; = \; 1$ and hence we deduce that $AM = AN$ . Thus the right-angled triangles $AMD$ and $AND$ are congruent, and hence $\angle FDA \,=\, \angle EDA$ , so the desired ratio is $\boxed{1}$ .