Answer:
Part 4) AA Similarity Postulate
Part 5)
The congruent angles are
m<ACB=m<CED , m<ABC=m<CDE, m<BAC=m<DCE
The proportional segments are
[tex]\frac{AB}{DC}=\frac{BC}{DE}=\frac{AC}{CE}[/tex]
Step-by-step explanation:
Part 4) we know that
If BC is parallel to DE
then
Triangles ABC and CDE are similar by AA Similarity Postulate ( the three interior angles are congruent)
so
m<ACB=m<CED ------> by corresponding angles
m<ABC=m<CDE -----> both angles measure is 90 degrees
m<BAC=m<DCE -----> the sum of the interior angles of a triangle must be equal to 180 degrees
If two figures are similar, then the ratio of its corresponding sides is equal
so
[tex]\frac{AB}{DC}=\frac{BC}{DE}=\frac{AC}{CE}[/tex]
substitute the values
[tex]\frac{6}{4}=\frac{8}{DE}=\frac{AC}{CE}[/tex]
Find DE
[tex]\frac{6}{4}=\frac{8}{DE}[/tex]
[tex]DE=8*4/6=16/3\ units[/tex]
In the triangle ABC
Applying Pythagoras Theorem
Find AC
[tex]AC^{2}=AB^{2}+BC^{2}[/tex]
[tex]AC^{2}=6^{2}+8^{2}[/tex]
[tex]AC^{2}=100[/tex]
[tex]AC=10\ units[/tex]
Find CE
[tex]\frac{6}{4}=\frac{AC}{CE}[/tex]
[tex]\frac{6}{4}=\frac{10}{CE}[/tex]
[tex]CE=10*4/6=20/3\ units[/tex]
