In this case, we'll have to carry out several steps to find the solution.
Step 01:
triangle diagram:
AC = 4
AB = 3
DC = 2
Step 02:
geometry:
similar triangles:
Triangle ABC similar to Triangle DEC:
AA (angle-angle):
∠CAB = ∠ CDE
∠ ABC = ∠ DEC
The triangles are similar
BC:
BC² = AC² + AB²
BC² = (4)² + (3)²
[tex]BC\text{ =}\sqrt{16+9}=\sqrt{25}=5[/tex]
BC = 5
DE:
[tex]\begin{gathered} \frac{AC}{DC}=\frac{AB}{DE} \\ \\ \frac{4}{2}=\frac{3}{DE} \\ \\ 4\text{ * }DE\text{ =3 * 2} \\ \\ DE\text{ = }\frac{6}{4}=\frac{3}{2}=1.5 \end{gathered}[/tex]
DE = 1.5
EC:
EC² = DC² + DE²
EC² = (2)² + (1.5)²
[tex]EC=\sqrt{4+2.25}=\sqrt{6.25}=2.5[/tex]
EC = 2.5
That is the full solution.
