In the trapezoid ABCD, ad ‖ BC, angle a = 90 °, ab = 7, ad = 2, BC = 3, whether there is a point P on the straight line AB, and find out the length of AP In the trapezoidal ABCD, ad ‖ BC, ∠ a = 90 °, ab = 7, ad = 2, BC = 3, ask whether there is a point P on the line AB, so that the triangle with P, a, D as the vertex is similar to the triangle with P, B, C as the vertex? If not, please explain the reason; if so, how many p points are there? And calculate the length of AP

In the trapezoid ABCD, ad ‖ BC, angle a = 90 °, ab = 7, ad = 2, BC = 3, whether there is a point P on the straight line AB, and find out the length of AP In the trapezoidal ABCD, ad ‖ BC, ∠ a = 90 °, ab = 7, ad = 2, BC = 3, ask whether there is a point P on the line AB, so that the triangle with P, a, D as the vertex is similar to the triangle with P, B, C as the vertex? If not, please explain the reason; if so, how many p points are there? And calculate the length of AP

Let AP = x, BP = 7-x
(1) From △ ADP to BCP,
∴AD/BC=x/（7-x）
x=14/5.∴AP=14/5.
（2）APD∽BCP,
∴x/3=2/（7-x）
x²-7x+6=0
（x-6）（x-1)=0
∴x=6,x=1
That is AP = 6, AP = 1, AP = 14 / 5, a total of three points

As shown in the figure, in ladder ABCD, AD / / BC, (AD

It is proved that if ad is parallel to BC, then ∠ APB = ∠ PBC;
So ⊿ APB ⊿ PBC, BP / BC = AP / BP, that is, BP square = BC times AP

If the slope I of a dam is 1:3 and the slope length AB is 20m, the height of the dam is ()
A. 10m B. 20m C. 40m D. 203m

As shown in the figure: ∵ slope I = 1:3, ∵ let AC = x, BC = 3x, according to Pythagorean theorem, ac2 + BC2 = AB2, then x2 + (3x) 2 = 202, the solution is x = 10