Points $d$, $e$, and $f$ are the midpoints of sides $\overline{bc}$, $\overline{ca}$, and $\overline{ab}$ of $\triangle abc$, respectively, and $\overline{cz}$ is an altitude of the triangle. if $\angle bac = 71^\circ$, $\angle abc = 39^\circ$, and $\angle bca = 70^\circ$, then what is $\angle ezd+\angle efd$ in degrees?

Consider ΔABC, DF is a midline of this triangle, so the midline joining the midpoints of two sides is parallel to the third side and half as long. You have that DF||AC, then ∠DFB=∠CAB=71°. Similarly, EF is a midline and EF||BC and therefore ∠AFE=∠ABC=39°. ∠AFE+∠EFD+∠DFB=180°, then 

∠EFD=180°-71°-39°=70°.

Consider right ΔAZC, ZE is a median and is equal to half of hypotenuse AC, then ΔAEZ is isoscales and ∠EAZ=∠EZA=71°. Similarly, ΔCZB is right, FD is a median and is equal to half of hypotenuse BC. ΔBDZ is isoscales and ∠DFZ=∠DBZ=39°.∠AZE+∠EZD+∠DZB=180°, then 

∠EZD=180°-71°-39°=70°.

Hence, ∠EZD+∠EFD=70°+70°=140°

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The answer to your question is d 68%