Answer:
Step-by-step explanation:
1) ΔCPD & ΔEPF
∠CPD = ∠EPF { Vertically opposite angles}
∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}
ΔCPD ≈ΔEPF {AA criteria for similarity }
[tex]\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\[/tex]
Cross multiply
EF * 15 = 27 * 7.5
[tex]EF =\frac{27*7.5}{15}\\\\[/tex]
EF = 27 * 0.5
EF = 13.5 cm
ii) EF // AB, so Triangles ACB & ECF are similar triangles
[tex]\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}[/tex]
[tex]AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm[/tex]
AC = 37.5 cm
