Question

На сторонах bc и cd квадрата abcd взяты точки h и m так, что bh:hc=cm:md=1:2. отрезки ah и bm пересекаются в точке p. найти отношения ap:ph.помогите пожалуйста с обьяснением и чертежом если понадобится

Answer 1

Проводим прямую параллельную  сторонам квадрата.(hL)
  прямоугольные  треугольники mBC и BLh подобны по общему острому углу.
А  треугольники BpA и phL подобны по  внутренним накрест лежащим углам. На рисунке обозначены  доли отношений. 
Ответ:9:1

Answer 2

Problem Analysis

We are given a square ABCD, and points H and M are taken on the sides BC and CD, respectively, such that BH:HC = CM:MD = 1:2. The segments AH and BM intersect at point P. We need to find the ratio AP:PH.

Solution

To find the ratio AP:PH, we can use similar triangles. Let's analyze the given information step by step.

1. We are given that BH:HC = 1:2. This means that the length of BH is one-third of BC, and the length of HC is two-thirds of BC. 2. Similarly, we are given that CM:MD = 1:2. This means that the length of CM is one-third of CD, and the length of MD is two-thirds of CD.

Now, let's consider the triangle ABH. Since BH:HC = 1:2, we can say that the length of AH is two-thirds of AB. Similarly, in triangle BMD, the length of BM is two-thirds of BD.

Since AB = BD (as ABCD is a square), we can conclude that the length of AH is two-thirds of AB, and the length of BM is two-thirds of AB.

Now, let's consider the triangle APH. We know that AH is two-thirds of AB, and BM is two-thirds of AB. Therefore, AP:PH = AH:BM = 2:2/3 = 6:2.

Hence, the ratio AP:PH is 6:2, which simplifies to 3:1.

Diagram

Here is a diagram to help visualize the problem:

      A
      |
      |
      |
B-----P-----H
|           |
|           |
|           |
M-----------C
      |
      |
      |
      D

In the diagram, AB = BC = CD = DA, and BH:HC = CM:MD = 1:2. The segments AH and BM intersect at point P.

I hope this explanation helps! Let me know if you have any further questions.