contestada

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?

Respuesta :

frika
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°
Ver imagen frika

Otras preguntas