∴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).