 \text{Je daná priamka }p=PQ\text{ a body } A,B: B∈\overrightarrow{pA}.  \
\text{Určte bod }X∈p: AX+BX \text{ bol čo najmenší.}

Postup

1.  \; {A': AA'⟂ \; p, \; d(A,p)=d(A',p)

2.  \; X: X=p∩BA'

Dôkaz 

2.  \; Y≠X, Y∈p

d(A',Y)+d(Y,B)>d(A',B) \text{ - troj. nerovnosť }_□

P = (2.6182727225147056, 3.193903586016193)
Q = (6.355286992499436, 4.333067787275481)
f = Line(P, Q)
A = (3.076716852289784, 4.527559236270969)
B = (5.4036681170572285, 4.840134779299432)
A' = Mirror(A, f)
g = Line(A', B)
X = Intersect(g, f)
h = Segment(A, X)
i = Segment(X, B)
s1: h + i
j = Segment(X, A')
text7: "3.  \; d(A,X)=d(A',X)=" + (LaTeX[h]) + ""
text8: "1. \; d(A,X)+d(B,X)= d(A',X)+d(B,X)=d(A',B)=" + (LaTeX[s1]) + ""
Y = Point(f)
k = Segment(A, Y)
m = Segment(Y, B)
l = Segment(Y, A')
s2: k + m
text9: "3.  \; d(A,Y)+d(B,Y)= d(A',Y)+d(Y,B)=" + (LaTeX[s2]) + ">" + (LaTeX[s1]) + "=d(A',B)"