As shown in the figure, the distance between a and B on the railway is 25km, C and D are two villages, Da ⊥ AB is in a, CB ⊥ AB is in B, known Da = 15km, CB = 10km, now we need to build a local product acquisition station E on the railway AB, so that the distance between C and D villages and E station is equal, then how many kilometers should e station be built from a station?

As shown in the figure, the distance between a and B on the railway is 25km, C and D are two villages, Da ⊥ AB is in a, CB ⊥ AB is in B, known Da = 15km, CB = 10km, now we need to build a local product acquisition station E on the railway AB, so that the distance between C and D villages and E station is equal, then how many kilometers should e station be built from a station?

Let AE = XKM, ∵ C and D villages have the same distance to e station, ∵ de = CE, that is, de2 = CE2. According to Pythagorean theorem, we can get 152 + x2 = 102 + (25-x) 2, x = 10. Therefore, e point should be built 10km away from a station

The distance between stations a and B on the railway is 25km, C and D are two villages, if Da = 10km, CB = 15km
Da ⊥ AB is at point a, CB ⊥ AB is at point B, Da = 15km, BC = 10km. Now we need to build a local product acquisition station E on Railway AB, so that the distance from village C and D to station e is equal, then how many kilometers should station e be built from station a? How to find out without Pythagorean theorem

Without Pythagorean theorem, we have to draw

As shown in the figure, on the straight railway, the distance between a and B is 25km, C and D are two villages, Da = 10km, CB = 15km, Da ⊥ AB is in a, CB ⊥ AB is in B. now we need to build a transfer station E on AB, so that the distance between C and D villages and E station is equal. How far should e be built from a?

Let AE = x, then be = 25-x. according to Pythagorean theorem, de2 = ad2 + AE2 = 102 + x2 in RT △ ade, CE2 = BC2 + be2 = 152 + (25-x) 2 in RT △ BCE. From the meaning of the title, we can see: de = CE, so: 102 + X2 = 152 + (25-x) 2, the solution is: x = 15km. (6 points). Therefore, e should be built 15km away from point a

As shown in the figure, the vertices D and C passing through ▱ ABCD make vertical lines de and CF respectively to the straight line where AB is on the opposite side, and the vertical feet are points E and f respectively. The verification is: AE = BF

It is proved that the ∵ quadrilateral ABCD is a parallelogram, ∵ ad = BC, ad ∥ BC, ∵ a = ∠ CBF, ∵ de ⊥ AB, CF ⊥ AB, ∵ DEA = ∠ f = 90 ° in △ AED and △ BFC, ≌ AE = BF