```Assignment #4
Nicole Mosteller
EMAT 6680
Assignment #4: Investigation #8.
Begin with acute DABC. Construct DMIG and DFDE (see instructions below).
Prove DMIG is congruent to DFDE.

Figure 1.```

To construct DMIG(see Figure 2): Let pt. M be the midpoint of segment BC, pt. I be the midpoint of segment AB, and pt. G be the midpoint of segment AC. By definition, DMIG is a medial triangle.

```
Figure 2.```
`How do the sides of DMIG relate to the sides of DABC?`

Because the vertices of DMIG are all midpoints of DABC, DMIG is a medial triangle whose sides are midsegments of DABC. Since the sides are midsegments, we can conclude:

```

.
Click here! to see why the midsegment is half the length of its corresponding side.```

To construct DFDE(see Figure 3): Construct the orthocenter (pt. H) of DABC. Click here! to see that DFDE can be constructed with pt. H anywhere inside the original triangle (not just at the orthocenter). To construct DFDE, first connect the vertices of DABC and the pt. H creating segments CH, BH, and AH. Let pt. D be the midpoint of segment CH, pt. E be the midpoint of segment BH, and pt. F be the midpoint of segment AH. When using the orthocenter, pt. H, as a vertex, DFDE is called a midsegment triangle.

`Figure 3.`

How do the sides of DFDE relate to the sides of DABC?

```To see this relation, it is necessary to use the smaller triangles that have the common vertex H
(DBCH, DABH, and DCAH. In DBCH, notice that pts. E and D are midpoints of two of its sides.
This means that segment ED is a midsegment of DBCH, and
.
In DABH, pts. F and E are midpoints of two of its sides.
By definition, segment EF is a midsegment of DABH, and
.
In DCAH, pts. F and D are midpoints of two of its sides.
By definition, segment FD is a midsegment of DCAH, and
.
```
`Let's compile the conclusions made on Figure 2 and Figure 3.`
```

Using the transitive property, we can now conclude :
MI = FD
IG = DE
GM = EF.
By SSS, DMIG is congruent to DFDE.