∴DH:CH:CD=5: 12: 13

∫CH = 60/ 13

∴DH=25/ 13,CD=65/ 13

∠∠DHC =∠DCB = 90°, ∠ CDH is the angle * * *,

∴△DHC∽△DCB

∴DH:CD=5: 13=CD:BD

∴bd=( 13/5)cd=( 13/5)×(65/ 13)= 13

(2) Core: Calculate the height and area of △BCE or △DEP by triangle similarity method, and then calculate the area by indirect method.

∫△DHC∽△DCB

∴cd:bc:bd=dh:ch:cd=5: 12: 13

∫BD = 13，

∴CD=5,BC= 12 and AB=5, AD= 12.

∫PD∨BC

∴∠DPE=∠BCE,∠CBE=∠PDE

∴△DEP∽△BEC

Let the height of △DEP be m and the height of △ BEC be n, then

M/n=DP/BC (triangles are similar, and the ratio of heights is equal to the similarity ratio)

AP = x，

∴DP= 12-x, which means m/n=( 12-x)/ 12.

∴m=( 12-x)/ 12×n[ 1]

∫m+n = CD = 5[2]

From [1][2], n=60/(24-x)

∴s△bec= 1/2×bc×n=360/(24-x)

∫S trapezoidal ABCP =1/2× (x+12 )× 5 = 5x/2+30.

∴S quadrilateral ABEP=S trapezoid ABCP-S△BEC=5x/2-360/(24-x)+30.

That is, y = 5x/2-360/(24-x)+30,0 < x <12.

(3) Core: Find the height m of △DEP by a method similar to △BEC.

∫ s △ Abd =1/2×12× 5 = 30, s quadrilateral ABEP=5S△DEP.

∴S△DEP=30/6=5

It can also be obtained from [1][2] in the problem (2).

m=(60-5x)/(24-x)

∫ 1/2×PD×m = 5

∴ 1/2×( 12-x)(60-5x)/(24-x)=5

∴x=6

∴AB=5,AP=6,DP=6,CD=5

Is a right triangle,

∴ Two triangles are similar, and the similarity ratio is 1 (two triangles are congruent).