Answer:
[tex]\text{AE} = \dfrac{50}{3}[/tex].
Step-by-step explanation:
Step 1: Show that triangle ABC is similar to triangle CDE.
Segment BC is parallel to segment DE. As a result,
[tex]\angle \text{BCD} = \angle \text{CDE} = 90\textdegree{}[/tex].
Point A, C, and E lines up on segment AE. As a result, [tex]\angle \text{ACE} = 180 \textdegree{}[/tex].
[tex]\angle \text{ACB} + \angle \text{BCD} + \angle \text{DCE} = \angle \text{ACE} = 180 \textdegree{}[/tex].
As seen above, [tex]\angle \text{BCD} = 90 \textdegree{}[/tex]. Therefore,
[tex]\angle \text{ACB} +90\textdegree{}+ \angle \text{DCE} = \angle \text{ACE} = 180 \textdegree{}\\\angle \text{ACB} +\angle \text{DCE} = 90 \textdegree{}[/tex].
The three angles of triangle ABC adds up to [tex]180 \textdegree{}[/tex]. That is:
[tex]\angle\text{BAC} + \angle \text{ACB}+ \angle \text{ABC} = 180 \textdegree{}[/tex].
[tex]\angle\text{ABC} = 90\textdegree{}[/tex] for being a right angle. As a result,
[tex]\angle\text{BAC} + \angle \text{ACB}+ 90 \textdegree{} = 180 \textdegree{}\\\angle\text{BAC} + \angle \text{ACB} = 90 \textdegree{}[/tex]
[tex]\angle\text{BAC} + \angle \text{ACB} = 90 \textdegree{} = \angle \text{ACB} +\angle \text{DCE} \\\angle\text{BAC} + \angle \text{ACB} = \angle \text{DCE} + \angle \text{ACB} \\\angle\text{BAC}= \angle \text{DCE}[/tex]
Two of triangle ABC and CDE's angles are the same, implying that the third angle is also the same. As a result, the two triangles are similar.
Step 2: Find the ratios between the segments.
Only the length of CD is known among all three sides in triangle CDE. Side CE is part of segment AE; the length of side CE needs to be determined. Side CD corresponds to side AB, whereas side CE corresponds to side AC.
[tex]\dfrac{\text{AC}}{\text{AB}} = \dfrac{\text{CE}}{\text{CD}}[/tex].
Step 3: Find the length of AC, CE, and hence AE.
Apply the Pythagorean Theorem:
[tex]\text{AC} = \sqrt{\text{AB}^{2} + \text{BC}^{2}}\\\phantom{\text{AC}} = \sqrt{6^2 + 8^2}\\\phantom{\text{AC}} = 10[/tex]
Apply the ratio found in step 2:
[tex]\text{AC} = \text{AB} \cdot \dfrac{\text{CE}}{\text{CD}}\\\phantom{\text{AC}} = 10 \times \dfrac{4}{6}\\\phantom{\text{AC}} = \dfrac{20}{3}[/tex].
[tex]\text{AE} = \text{AC} + \text{CE} = 10 + \dfrac{20}{3} = \dfrac{50}{3}[/tex].
