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

Are you in Aops geometry?

Answer 2

She is defintley in Aops geometry. I have the same problem due this week.

Answer 3

we are all cheaters haha

Answer 4

GIUVE NA ANSwer

Answer 5

[AI-generated answer — Explain Bot]

To find the value of \(\angle EZD + \angle EFD\), we first need to determine the values of \(\angle EZD\) and \(\angle EFD\).

Let's start by drawing a diagram of \(\triangle ABC\) and the given information.

Since \(D, E, F\) are midpoints of the sides \(\overline{BC}, \overline{CA}, \overline{AB}\) respectively, we can draw the line segments \(DE, EF, FD\) to form the smaller triangles within \(\triangle ABC\).

Given that \(\overline{CZ}\) is an altitude of \(\triangle ABC\), we can label the intersection point of \(CZ\) and \(DE\) as point \(X\) and the intersection of \(CZ\) and \(FD\) as point \(Y\).

Now, we can see that \(\triangle DXZ\) and \(\triangle CYZ\) are similar to \(\triangle ABC\) by angle-angle similarity.

Using the given angle measurements, we can determine the values of \(\angle DXZ\) and \(\angle CYZ\).

\(\angle BAC = 71^\circ\), so \(\angle CZY = 90^\circ - 71^\circ = 19^\circ\).

Since \(\angle CZY = \angle CYZ\) in \(\triangle CYZ\), we have \(\angle CYZ = 19^\circ\).

Similarly, \(\angle ABC = 39^\circ\), so \(\angle DXZ = \angle DZB = 39^\circ\).

Now, let's find the values of \(\angle EZD\) and \(\angle EFD\). These angles are the internal angles of the triangles formed by the line segments \(DE, EF, FD\).

Considering \(\triangle DXE\), we know that the sum of the angles in a triangle is \(180^\circ\).

So, \( \angle DXE + \angle EXD + \angle EDE = 180^\circ\).

Since \( \angle DXE = \angle DXZ\) and \(\angle EXD = \angle EZD\), we have \(39^\circ + \angle EZD + 90^\circ = 180^\circ\).

Simplifying further, we get \( \angle EZD = 180^\circ - 39^\circ - 90^\circ = 51^\circ\).

Similarly, considering \(\triangle DFE\), we can apply the same reasoning to find that \( \angle EFD = \angle DFE = 39^\circ\).

Finally, we can calculate the sum of \(\angle EZD + \angle EFD\).

\(\angle EZD + \angle EFD = 51^\circ + 39^\circ = 90^\circ\).

Therefore, \(\angle EZD + \angle EFD = 90^\circ\).

Answer 6

Stop cheating!