As shown in the figure, P is a point in the square ABCD, and there is a point e outside the square ABCD, which satisfies ∠ Abe = ∠ CBP, be = BP,

As shown in the figure, P is a point in the square ABCD, and there is a point e outside the square ABCD, which satisfies ∠ Abe = ∠ CBP, be = BP,

Solution: connecting PE
In △ Abe and △ CBP,
BE=BP,∠ABE=∠CBP,AB=BC
∴△ABE≌△CBP(SAS)
The quadrilateral ABCD is a square
∴∠ABC=∠ABP+∠CBP=90°
∴∠ABE+∠ABP=∠ABP+∠CBP=90°
∴∠EBP=∠ABC=90°
A BPE is an isosceles right triangle
∴∠BPE=45°
∵∠APB=135°
∴∠APE=90°
In △ BPE, sin ∠ BPE = sin45 ° = be ∶ PE = 1: √ 2
∴PE=BE÷sin45°=√2BE
∵BE=BP,PA=2PB
∴PA=2BE
In RT △ ape, according to Pythagorean theorem
AE2 = ap2 + Pe2 =? Be2 + 2be2 = three thirds be
∴AP=?PB=?BE
Ψ AP: AE = 1 / 2 be: 3 / 2 be
That is, one-half: three-thirds = 1:3