如图,A,B是公路l（l为东西走向）两旁的两个村庄,

设定直线AB交叉直线L于O,
既然B村在A村的南偏东45°方向上,
AC = CO ；BD = DO
CO = 1 ；DO = 2
AO = √(AC² + CO²)
= √(1 + 1)
= √2
BO = √(BD² + DO²)
= √(2 + 2)
= 2
AB的距离为 AO + BO
= (√2) + 2
= 1。4142 + 2
= 3。4142 km
P点是AB的中间,也就是,
AP = BP = (2 + √2)/2 = 1。7071
拉长AC的直线,并从P点向西画条直线交叉拉长过后的AC直线,并交叉于Q。
AQ的长度是,
AP = √(AQ² + PQ²) ,AQ = PQ
AP² = 2AQ²
AQ = AP/√2
AQ = 1。4142 = PQ
CQ = AQ - AC
= 1。4142 - 1
= 0。4142
CP = √(CQ² + PQ²)
= √(0。1716 + 2)
= √2。1716
= 1。4736km