Question

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?

Answer 1

The sum of $\boxed{\angle {\text{EZD}} + \angle \text{EFD} = {{140}^ \circ }}$.

Further Explanation:

The sum of all angle of the triangle is $\boxed{{{180}^ \circ }}$.

The line joining the mid points of a triangle is parallel to the third side and half of it.

Given:

D, E and F are the midpoints of the sides BC, CA and AB respectively.

CZ is an altitude of the triangle ABC.

$\angle \text{BAC} = {71^ \circ }$, $\angle \text{ABC} = {39^ \circ }$ and $\angle \text{BCA} = {70^ \circ }$.

Calculation:

E is the midpoint of AC and D is the midpoint of BC.

The line ED is parallel to the line AB.

F is the midpoint of AB and D is the midpoint of AB.

The line FD is parallel to the line AC.

E is the midpoint of AC and F is the midpoint of AB.

The line EF is parallel to the line BC.

Therefore, $\angle \text{AFE} = \angle \text{ABC} = 39^\circ$.

$\angle \text{AFE} + \angle \text{EFD} + \angle \text{DFB} = 180^\circ$

Substitute $39^\circ$ for $\angle \text{AFE}$ and $71^\circ$ for $\angle \text{DFB}$ in the above equation to obtain the value of angle $\angle \text{EFD}$.

$\begin{aligned}\angle EFD &= {180^ \circ } - {71^ \circ } - {39^ \circ } \\ &= {180^ \circ } - {110^ \circ } \\ &= {70^ \circ } \\\end{aligned}$

In triangle AEZ the side AE is equal to side EZ.

Angle opposite to equal side are equal Therefore, $\angle \text{EAZ} = \angle \text{EZA} = {71^ \circ }$ and $\angle \text{DFZ} = \angle \text{DBZ} = {39^ \circ }$.

The sum of $\angle \text{AZE} + \angle \text{EZD} + \angle \text{DZB} = {180^ \circ }$.

Substitute $39^\circ$ for $\angle \text{DZB}$ and $71^\circ$ for $\angle \text{AZE}$ in the above equation to obtain the value of $\angle \text{EZD}$.

$\begin{aligned}\angle \text{EZD}+{71^ \circ }+ {39^ \circ } &= {180^\circ } \\\angle \text{EZD} &= {180^\circ } - {39^ \circ } - {71^ \circ } \\\angle \text{EZD} &= {180^\circ } - {110^ \circ } \\\angle \text{EZD} &= {70^ \circ } \\\end{aligned}$

The sum of $\angle {\text{EZD}} + \angle \text{EFD}$ can be calculated as,

$\begin{aligned}\angle {\text{EZD}} + \angle \text{EFD} &= {70^ \circ } + {70^ \circ } \\&= {140^ \circ } \\\end{aligned}$

Kindly refer to the image attached.

The sum of $\boxed{\angle {\text{EZD}} + \angle \text{EFD} = {{140}^ \circ }}$.

Learn more:

1. Learn more about rotation of triangle brainly.com/question/2992432

2. Learn more about angles brainly.com/question/1953744

3. Learn more about coordinates of triangle brainly.com/question/7437053

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Triangle

Keywords: sum, angle, altitude, triangle, parallel degree, midpoints, sides, points, triangle ABC, midline, third side, median, hypotenuse, $\angle \text{BAC} = {71^ \circ }$, $\angle \text{ABC} = {39^ \circ }$, $\angle \text{BCA} = {70^ \circ }$.

Answer 2

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°