As shown in the figure, in △ ABC, D, e and F are the points on AB, AC and BC respectively, and de ‖ BC, EF ‖ AB, ad: DB = 2:3, BC = 20cm, the length of BF can be obtained

The length of BF is 8cm because of the fact that bdef is parallelogram, BF = De, ad: DB = 2:3, ad: ab = 2:5, de: BC = ad: ab = 2:5, that is, de: 20 = 2:5, de = 8, BF = 8

As shown in the figure, in △ ABC, D, e and F are the points on AB, AC and BC respectively, and de ‖ BC, EF ‖ AB, ad: DB = 2:3, BC = 20cm, the length of BF can be obtained

The length of BF is 8cm because of the fact that bdef is parallelogram, BF = De, ad: DB = 2:3, ad: ab = 2:5, de: BC = ad: ab = 2:5, that is, de: 20 = 2:5, de = 8, BF = 8

It is known that as shown in the figure, triangle ABC and triangle def are right triangles, AC is vertical BF, De is vertical BF, AC = BC = 8, de = EF = 6
Translation. Find 1 when the vertex D of triangle DEF is translated to the edge ab of triangle ABC, what is the overlapping area? 2 when the vertex e of triangle DEF is translated to the midpoint of BC, what is the overlapping area?

It is known that as shown in the figure, triangle ABC and triangle def are right triangles, AC vertical BF, de vertical BF, AC = BC = 8, de = EF = 6. Translate the triangle def along CB direction. Find 1. When the vertex D of triangle DEF is translated to the edge ab of triangle ABC, what is the overlapping area? 2. When the top point e of triangle DEF is translated to the midpoint of BC, what is the overlapping area?
1. When the vertex D of the triangle DEF is translated to the edge ab of the triangle ABC, DF and AC intersect at M,
Triangle DEB is an isosceles right triangle
BE=DE=6
CE=BC-BE=8-6=2
CF=EF-CE=6-2=4
The triangle MCF is an isosceles right triangle
The overlapping area is trapezoid mced = triangle DFE - Triangle MCF = 6 * 6 / 2-4 * 4 / 2 = 10
When the vertex e of the triangle DEF is translated to the midpoint of BC, DF intersects AC at m, AB at n, and de intersects AB at mo
The triangle OEB is an isosceles right triangle
OE=BE=BC/2=4=CE
Triangle FCM is isosceles right triangle
CM=FC=FE-CE=6-4=2
The triangle amn is an isosceles right triangle
Bevel am = ac-cm = 8-2 = 6
AN^2+MN^2=36
AN^2=18
Overlap area = triangle ABC triangle OBE triangle amn = 8 * 8 / 2-4 * 4 / 2-18 / 2 = 32-8-9 = 15

As shown in the figure, in △ ABC, EF ‖ CD, de ‖ BC

∵ EF ∥ CD, de ∥ BC, ∥ AFFD = AEEC, addb = AEEC, ∥ AFFD = addb, namely AF: FD = ad: dB

As shown in the figure, in △ ABC, D, e and F are the points on AB, AC and BC respectively, and de ‖ BC, EF ‖ AB, ad: DB = 2:3, BC = 20cm, the length of BF can be obtained