solve this---**PROBLEMS**
**1-11** In any $\triangle ABC$, $D, E,$ and $F$ are midpoints of the sides $\overline{AC}, \overline{AB},$ and $\overline{BC},$ respectively (Fig. 1-11). $\overline{BG}$ is an altitude of $\triangle ABC.$ Prove that $\angle EGF \cong \angle EDF.$
**Challenge 1** Investigate the case when $\triangle ABC$ is equilateral.
**Challenge 2** Investigate the case when $AC = CB.$
**Diagram 1-11 Description:**
* Type: Geometric figure (Triangle).
* Main Elements:
* A triangle labeled $\triangle ABC$ with vertices A, B, and C.
* Point A is on the left, B on the right, and C is above the segment AB.
* Side AC has a point D marked as the midpoint.
* Side AB has a point E marked as the midpoint.
* Side BC has a point F marked as the midpoint.
* Segments DE, EF, and DF are drawn, forming the medial triangle $\triangle DEF$.
* A segment BG is drawn from vertex B to side AC, perpendicular to AC. G is the foot of the altitude on AC. The segment BG is labeled as an altitude. A right angle symbol is shown at G, indicating that BG is perpendicular to AC.
* A segment EG is drawn.
* A segment FG is drawn.
**Diagram 1-12 Description:**
* Type: Geometric figure (Triangle).
* Main Elements:
* A triangle labeled $\triangle ABC$ with vertices A, B, and C.
* Vertex C is at the top left, Vertex A is at the bottom left, and Vertex B is at the bottom right.
* A right angle symbol is shown at vertex C, indicating $\angle ACB = 90^\circ$.
* Point G is on side AC. A segment BG is drawn from B to G. A right angle symbol is shown at G on AC, indicating that BG is perpendicular to AC, making BG an altitude to AC.
* Point D is on side AB. A segment CD is drawn from C to D. A right angle symbol is shown at D on AB, indicating that CD is perpendicular to AB, making CD an altitude to AB.
* Point F is on side BC. A segment AF is drawn from A to F. A right angle symbol is shown at F on BC, indicating that AF is perpendicular to BC, making AF an altitude to BC.
* Point E is on side AB. A segment CE is drawn from C to E. A right angle symbol is shown at E on AB, indicating that CE is perpendicular to AB, making CE an altitude to AB. (Note: There are two altitudes shown originating from C and A, and their feet are labeled D, E, and F. It seems D and E are both on AB, and F is on BC. There might be a labeling inconsistency in the diagram or the problem it is associated with, but the elements are as described).
* The diagram shows the intersection of altitudes. BG, CD, and AF appear to be altitudes. However, the labels D, E, and F are placed on the sides AB and BC, which are the feet of altitudes from vertices C, A, and A respectively. The diagram shows CD $\perp$ AB, AF $\perp$ BC, and BG $\perp$ AC. The points D and E are labeled on AB, and F is labeled on BC. The segment CE is also drawn from C to E on AB, with a right angle at E.
* The diagram seems to illustrate the concept of altitudes in a right-angled triangle.
*(Note: Problem 1-11 specifically refers to Fig. 1-11, which shows midpoints and one altitude in a general triangle. Fig. 1-12 is a separate diagram, likely related to a different problem, possibly 1-12 as labeled beside it, illustrating altitudes in a right triangle. The detailed description of 1-12 is provided as requested by the requirement to extract all content related to the question(s) from the image, but it is not directly related to Problem 1-11 itself.)*
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We have triangle ABC where D, E, and F are midpoints of sides AC, AB, and BC respectively. BG is an altitude from B to side AC, making angle BGA a right angle. We need to prove that angle EGF is congruent to angle EDF.
Now we analyze the key properties. Since BG is perpendicular to AC, we have two right triangles: triangle BGA and triangle BGC. In right triangle BGA, E is the midpoint of hypotenuse AB, so by the median to hypotenuse property, EG equals half of AB. Similarly, in right triangle BGC, F is the midpoint of hypotenuse BC, so FG equals half of BC.
Now we examine the medial triangle DEF formed by connecting the midpoints. By the Midpoint Theorem, each side of the medial triangle is parallel to and half the length of the corresponding side of the original triangle. Specifically, DE equals half of BC, EF equals half of AC, and DF equals half of AB.
Now we compare the side lengths of triangles EGF and EDF. From our previous analysis, EG equals half of AB, which equals DF from the medial triangle. Similarly, GF equals half of BC, which equals DE. The side EF is common to both triangles. Since all three pairs of corresponding sides are equal, the triangles are congruent by the Side-Side-Side congruence criterion.
We have successfully proven that angle EGF is congruent to angle EDF. The proof relied on the median to hypotenuse property in right triangles, the midpoint theorem for the medial triangle, and the Side-Side-Side congruence criterion. This demonstrates how geometric properties work together to establish angle relationships in complex figures.