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\small{\mbox{Consider two points }}\definecolor{azul}{rgb}{0,0.6,0.8} \textcolor{laranja}{A} \mbox{ and } \definecolor{laranja}{rgb}{1,0.6,0}  \textcolor{azul}{B} \mbox{ corresponding respectively to the positions of} 
\ \small\mbox{the lifeguard and the swimmer.}
\ \small\mbox{\textbf{We want to show } that the fastest path of the lifeguard is the path A-C-B}
\ \small\mbox{such that: } \ \;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\frac{\;\;\frac{\mbox{ } \overline {DC}\mbox{ }}{\overline {AC}}\;\;}{\frac{\mbox{ }\overline{EC}\mbox{ }}{\overline{BC}}}=\frac{\mbox{running speed}}{\mbox{swimming speed}}.
\
\scriptsize \mbox{(If you know about trigonometric functions, note that this equality is equivalent to }\ \scriptsize {\frac{\mbox{Sin }\beta}{\mbox{Sin }\alpha}}=\frac{\mbox{running speed}}{\mbox{swimming speed}}.)

\small\mbox{So consider another \textcolor{Red}{path} A-F-B, where \textcolor{Red}{F} is an arbitrary point of the line}
\ \small\mbox{segment DE, distinct from \definecolor{purpura}{rgb}{0.5,0,0.5}\textcolor{purpura}{C}. We will show that the lifeguard takes more}
\ \small \mbox{time to make it than to make A-C-B.}

\textcolor{Red}{\mbox{F}}

\small\mbox{Let us start by assuming that \textcolor{Red}{F} is in CE.}
\ \small\mbox{Let G be the orthogonal projection of \textcolor{Red}{F} over BC. }

\small \mbox{And let r be the perpendicular to AC  by \textcolor{purpura}{C}, let H be the orthogonal} \
\small\mbox{projection of \textcolor{Red}{F} over r and let I be the intersection of r with AF.} 

\small\mbox{Each one of paths A-C-B and A-F-B may be decomposed in 3 parts:}
\ \small\mbox{AC, CG and GB, AI, IF and FB, respectively. }
\ \small \overline{AI}>\overline{AC} \small \mbox{ and }  \overline{FB}>\overline{GB} \mbox{, since AI and FB are the hypotenuses of triangles}
\
\small \mbox{ACI and FGB and AC and GB are catheti of those triangles. Therefore,  \he lifeguard makes AC faster than AI and makes GB faster than FB. \It remains to check that he swims GB faster than he runs IF. }

\definecolor{verde2}{rgb}{0,.8,0}
\small \mbox{Let us start by comparing the times of CG and HF.} \ \small \mbox{The \textcolor{verde2}{\mbox{triangles ADC and CHF}} are similar, since }A \hat {D} C=C\hat H F=90\degree \ \mbox{ and } D\hat A C = H \hat CF=\beta.
 \small\mbox{ Then, } \frac{\mbox{ }\overline {DC}\mbox{ }}{\overline {AC}}=\frac{\mbox{ }\overline {HF}\mbox{ }}{\overline {CF}}.

\small \mbox{\textcolor{verde2}{\mbox{Triangles CEB and CGF}} are similar, since } C \hat {E} B=C\hat G F=90\degree \mbox{ and } \ \small E\hat C B =F \hat CG.
 \small\mbox{ Then, } \frac{\mbox{ }\overline {CE}\mbox{ }}{\overline {BC}}=\frac{\mbox{ }\overline {CG}\mbox{ }}{\overline {CF}}.

\small \mbox{According to the preceding identities, } \frac{\mbox{ }\overline {DC}\mbox{ }}{\overline {AC}}=\frac{\mbox{ }\overline {HF}\mbox{ }}{\overline {CF}} \mbox{ and } \frac{\mbox{ }\overline {CE}\mbox{ }}{\overline {BC}}=\frac{\mbox{ }\overline {CG}\mbox{ }}{\overline {CF}} \mbox{, the }
\ \small \mbox{identity } \frac{\;\;\frac{\mbox{ } \overline {DC}\mbox{ }}{\overline {AC}}\;\;}{\frac{\mbox{ }\overline{CE}\mbox{ }}{\overline{BC}}}=\frac{\mbox{running speed}}{\mbox{swimming speed}}
\small \mbox{, that determined point \textcolor{purpura}{C}, may be written } \
\small \mbox{as } \frac{\mbox{ }\overline{HF}\mbox{ }}{\overline{CG}}=\frac{\mbox{running speed}}{\mbox{swimming speed}} \mbox{ or, equivalently, } \frac{\mbox{ }\overline{HF}\mbox{ }}{\mbox{running speed}}=\frac{\overline{CG}}{\mbox{swimming speed}}.
\
\small \mbox{Thereby, the lifeguard spends the same time to run HF and to swim CG}.

\small \mbox{But }\overline{IF}>\overline{HF} \small \mbox{ since IF is the hypotenuse of triangle IHF and HF is a cathetus }\ \small \mbox{and, therefore, the lifeguard takes less time to make HF than IF. }
\ \small\mbox{We saw that the running time of HF and the swimming time of CG are equal. \Hence the lifeguard takes more time to run IF than to swim CG, as we}
\ \small  \mbox{wanted to show (see point 5).}
\
\small \mbox{In conclusion, if \textcolor{Red}{F} is a point in CE (distinct from \textcolor{purpura}{C}) the lifeguard spends }
\ \small \mbox{more time to make A-F-B than to make A-C-B. }

\textcolor{Red}{\mbox{F}}

\small \mbox{Suppose now that \textcolor{Red}{F} is in DC.}
\ \small \mbox{Let J be the orthogonal projection of \textcolor{Red}{F} over AC, let s be the perpendicular }

\ \small \mbox{to BC by \textcolor{purpura}{C}, let K be the orthogonal projection of \textcolor{Red}{F} over s and let L be the} \
\small\mbox{intersection of s with BF.} 

\small \mbox{With similar arguments, one may prove that the lifeguard takes more time}
\
\small \mbox{to run AF than AJ and takes the same time to swim FK and to run JC,}
\ \small \mbox{hence he takes more time to swim FL than to run JC and he takes more}
\ \small \mbox{time to swim LB than CB. \
Details are left to the reader.}

\mbox{r}

\definecolor{laranja}{rgb}{1,0.6,0}  \definecolor{azul}{rgb}{0,0.6,0.8}  \large\textbf{Move \textcolor{laranja}{A} or \textcolor{azul}{B}. For each choice, \definecolor{purpura}{rgb}{0.5,0,0.5}\textcolor{purpura}{C}  is the point that minimizes the time duration of path A-C-B.}\
\mbox{ }\ \mbox{We present a condition that determines \textcolor{purpura}{C} and we give a geometric argument for it.}

Forward

i = 0 (slider from 0 to 11, step 1)
spd_{running} = 4 (slider from 1 to 10, step 0.1)
spd_{swimming} = 2.5 (slider from 1 to 10, step 0.1)
f(y) = y > 0.01
A = PointIn(f)
D: (x(A), 0)
e(y) = y < -0.01
B = PointIn(e)
C_2 = Extremum(((x(B) - x(A) - x)^(2) + y(B)^(2))^(0.5) / spd_{swimming} + (x^(2) + y(A)^(2))^(0.5) / spd_{running}, -100, 100)
C: (x(A) + x(C_2), 0)
m = Segment(C, D)
E: (x(B), 0)
q = Segment(E, D)
b = Segment(A, B)
c = Segment(B, C)
d = Segment(C, A)
distânciaAB = Distance(A, B)
TextoAB: "\overline{" + (Name[A]) + (Name[B]) + "} \, = \, " + distânciaAB
distânciaAD = Distance(A, C)
TextoAD: "\overline{" + (Name[A]) + (Name[C]) + "} \, = \, " + distânciaAD
distânciaDB = Distance(C, B)
TextoDB: "\overline{" + (Name[C]) + (Name[B]) + "} \, = \, " + distânciaDB
E_2 = Intersect(a, b)
dist: distânciaAD + distânciaDB
distânciaAE = Distance(A, E_2)
TextoAE: "\overline{" + (Name[A]) + (Name[E_2]) + "} \, = \, " + distânciaAE
distânciaEB = Distance(E_2, B)
TextoEB: "\overline{" + (Name[E_2]) + (Name[B]) + "} \, = \, " + distânciaEB
tempmin: distânciaDB / spd_{swimming} + distânciaAD / spd_{running}
tempo = If((y(A) * y(B)) < 0, distânciaEB / spd_{swimming} + distânciaAE / spd_{running}, If[y(A) ≤ 0 ∧ y(B) ≤ 0, distânciaAB / spd_{swimming}, distânciaAB / spd_{running}])
g = OrthogonalLine(C, a)
F_2 = Point(g)
G_2 = Point(g)
β = If(x(B) > x(A), Angle[F_2, C, A], Angle[A, C, F_2])
α = If(x(B) > x(A), Angle[G_2, C, B], Angle[B, C, G_2])
rvel: spd_{running} / spd_{swimming}
rang: sin(β) / sin(α)
H_2 = Point(g)
h = Circle(C, H_2)
I_2 = Intersect(h, g, 2)
i_1 = Segment(H_2, I_2)
j = Segment(A, B)
n = Segment(E, B)
p = Segment(D, A)
F = Point(q)
k = Segment(A, F)
l = Segment(F, B)
r_1 = Segment(C, E)
D_1 = Point(r_1)
s_1 = Segment(A, D_1)
t = Segment(D_1, B)
a_1 = OrthogonalLine(D_1, c)
G = Intersect(a_1, c)
r = OrthogonalLine(C, d)
c_1 = OrthogonalLine(D_1, r)
H = Intersect(r, c_1)
I = Intersect(s_1, r)
polígono1, c_2, b_1, e_1 = Polygon(A, D, C)
polígono2, d_1, c_3, h_1 = Polygon(C, H, D_1)
polígono3, m_1, p_1, q_1 = Polygon(C, E, B)
polígono4, g_1, c_5, d_3 = Polygon(C, D_1, G)
J_1 = (9.11127596204854, 17.316482214911172)
f_1 = Segment(C, D)
H_1 = Point(f_1)
j_1 = Segment(A, H_1)
k_1 = Segment(H_1, B)
K_1 = Point(i_1)
l_1 = OrthogonalLine(H_1, d)
J = Intersect(l_1, d)
s = OrthogonalLine(C, c)
n_1 = OrthogonalLine(H_1, s)
K = Intersect(n_1, s)
L = Intersect(s, k_1)
I_1: (x(B) - 7, y(B) - 1)
L_1: (x(A) - 4, y(A) - 2)
N_1: (x(A) + 2.5, y(A))
P_1: (x(A) + 4, y(A) - 2)
O_1: (x(A) - 2.5, y(A))
γ = Angle(B, G, D_1)
Q_1 = Point(r)
R_1 = Point(c_1)
δ = Angle(Q_1, H, R_1)
ε = Angle(A, J, H_1)
ζ = Angle(L, K, H_1)
M_1: (x(B) - 0.7, y(B) - 1)
S_1 = Point(r)
texto8: "\small \frac{\;\;\frac{\mbox{ } \overline {DC}\mbox{ }}{\overline {AC}}\;\;}{\frac{\mbox{ }\overline{EC}\mbox{ }}{\overline{BC}}}=" + (LaTeX[rang]) + ""
texto7: "\small{\frac{\mbox{running speed}}{\mbox{swimming speed}}}=" + (LaTeX[rvel]) + " "