Question

На диагонали AC квадрата ABCD взята точка M так, что AM=AB. Через точку M проведена прямая, перпендикулярная к прямой AC и пересекающая BC в точке H. Докажите, что BH=HM=MC. Решить полностью с пояснениями, срочно!

Answer 1

Given Information

We are given a square ABCD, and a point M on the diagonal AC such that AM = AB. A line is drawn through point M, perpendicular to line AC, and it intersects BC at point H. We need to prove that BH = HM = MC.

Proof

To prove that BH = HM = MC, we will use the given information and properties of squares.

Step 1: Draw a diagram to visualize the given information. Let's label the points as shown below:

     A _______ M _______ C
      |                |
      |                |
      |                |
      |                |
      |                |
      B _______ H _______ D

Step 2: Since AM = AB, we can conclude that triangle ABM is an isosceles right triangle. This is because in a square, all sides are equal, and the diagonal divides the square into two congruent right triangles.

Step 3: In an isosceles right triangle, the altitude drawn from the right angle to the hypotenuse bisects the hypotenuse. Therefore, HM = MC.

Step 4: Since HM = MC, we can conclude that triangle HMC is an isosceles triangle.

Step 5: In an isosceles triangle, the altitude drawn from the vertex angle to the base bisects the base. Therefore, BH = HM = MC.

Hence, we have proved that BH = HM = MC.

Conclusion

In the given square ABCD, with point M on the diagonal AC such that AM = AB, and a line drawn through point M perpendicular to line AC intersecting BC at point H, we have proved that BH = HM = MC.

Note: The sources provided did not contain specific information related to the given problem. Therefore, the proof is based on the properties of squares and isosceles triangles.