The minimum sum of line segments in plane geometry is one of the most difficult and difficult problems for junior middle school students. Here, I have a look at the relevant senior high school entrance examination questions in some provinces and cities in China in 2005. This paper will analyze the test questions of senior high school entrance examination for reference.

First, take the square as the carrier to find the minimum value of the sum of line segments.

Example 1. As shown in figure 1, the quadrilateral ABCD is a square with a side length of 4, e is a point on BC, CE = 1, and p is any point on the diagonal BD, so the minimum value of PE+PC is _ _ _ _ _ _ _ _.

Analysis: Because BD is the diagonal of square ABCD and connects AP, it is easy to prove △ ADP △ CDP, so PA = PC. At this time, finding the minimum value of PE+PC is transformed into finding the minimum value of PA+PE and connecting AE. In △ PA+PE, because PA+PE is AE, when point P is the intersection of A and BD (that is, when points A, P and E are three)

Solution: the connection PA, ∫BD is the diagonal of the square ABCD.

∴AD=CD,∠ADP=∠CDP

And DP = DP, ∴△ADP≌△CDP.

∴PA=PC

Connecting AE

∵CE= 1,∴BE=3

In Rt△ABE,

According to the fact that the sum of two sides in a triangle is greater than the third side, when P is the intersection of AE and BD, the minimum value of PA+PE is AE, that is, PA+PE ≥ AE, ∴ PA+PE ≥ 5, that is, PE+PC ≥ 5, and the minimum value of ∴ PE+PC is 5 (only when the straight lines of A, P and E are * * *.