As shown in the figure, P is a point within square ABCD. If PA = 1, PB = 2, PC = 3, find the degree of ∠APB. Analysis: Because quadrilateral ABCD is a square, both sides are equal and perpendicular. We can rotate △ABP 90° clockwise around point B to get △CBE, which is connected t

As shown in the figure, P is a point within square ABCD. If PA = 1, PB = 2, PC = 3, find the degree of ∠APB.

Analysis: Because quadrilateral ABCD is a square,

so both sides are equal and vertical.

We can rotate △ABP 90° clockwise around point B to get △CBE, which is connected to PE.

can be obtained from the properties of rotation:

∠CEB =∠APB,

EB=PB,

EC=PA,

∠PBE=90° (because it is rotated 90°),

so △BEP For isosceles right triangle ,

so ∠BEP=45°.

In Rt△PBE,

because PB=2,

so PE=√PB²+EB²=2√2.

In △PEC,

because PC=3,

PE=2√2,

EC=PA=1,

so PC²=P E²+EC²

so ∠PEC=90°,

so ∠APB=∠CEB

=∠BEP+∠PEC

=45°+90°

=135°.

Summary: For a point in the square, you can rotate it 90° so that the three line segments are concentrated into the same quadrilateral (such as the quadrilateral PBEC in this question). After making the diagonal of the quadrilateral (such as PE in this question), you can find that one of the triangles (such as △PBE in this question) is an isosceles right triangle, so it can be concluded that PE=√2BP. In this way, √2BP, PC, and PA are concentrated into the same △PEC, which is indeed very strange.

This is exactly what the saying goes: make a difference and unify three line segments that are not in the same triangle into the same triangle.