Answer:
CE = 17
Step-by-step explanation:
∵ m∠D = 90
∵ DK ⊥ CE
∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK
In the two Δ KDE and KCD:
∵ m∠KDE = m∠KCD
∵ m∠DKE = m∠CKD
∵ DK is a common side
∴ Δ KDE is similar to ΔKCD
∴ [tex]\frac{KD}{KC}=\frac{DE}{CD}=\frac{KE}{KD}[/tex]
∵ DE : CD = 5 : 3
∴ [tex]\frac{KD}{KC}=\frac{5}{3}[/tex]
∴ KD = 5/3 KC
∵ KE = KC + 8
∵ [tex]\frac{KE}{KD}=\frac{5}{3}[/tex]
∴ [tex]\frac{KC+8}{\frac{5}{3}KC }=\frac{5}{3}[/tex]
∴ [tex]KC + 8 = \frac{25}{9}KC[/tex]
∴ [tex]\frac{25}{9}KC - KC=8[/tex]
∴ [tex]\frac{16}{9}KC=8[/tex]
∴ KC = (8 × 9) ÷ 16 = 4.5
∴ KE = 8 + 4.5 = 12.5
∴ CE = 12.5 + 4.5 = 17
