Cho tam giác ABC vuông tại A có AB < AC, gọi M là trung điểm của AB

To solve the problems in the given geometry exercises, we can break them down systematically.

### Problem 32
Given a right triangle \( ABC \) with a right angle at \( A \), and \( AB < AC \).

1. **a)** To show \( AM = \frac{1}{2} AB \):
- Since \( M \) is the midpoint of \( AB \), by the definition of a midpoint, we have \( AM = MB = \frac{AB}{2} \).

2. **b)** To prove that triangle \( DMCE \) is a right triangle:
- Here, you can use the properties of right triangles and perpendicular lines. Since \( D \) and \( E \) lie on the extensions or perpendiculars from points \( A \) and \( C \), you can apply the Pythagorean theorem or consider the angles formed to show the right triangle.

3. **c)** To show \( DH \perp AC \):
- Since \( H \) lies on \( BC \) and is connected to \( AC \), you can demonstrate that the angle between line segments \( DH \) and \( AC \) is \( 90^\circ \).

### Problem 33
Given the square \( ABCD \) with \( \angle D = 90^\circ \) and \( AB = AD = \frac{1}{2} CD \):

1. **a)** To show that quadrilateral \( ABHD \) is a square:
- You can illustrate that all sides are equal and that all angles are \( 90^\circ \). This follows from the properties of squares and the fact that \( AB \) and \( AD \) are both equal and meet at a right angle.

2. **b)** To demonstrate \( M \) as the midpoint:
- By definition, \( M \) is halfway between \( B \) and \( H \). Using distance formulas, you can calculate and verify that \( AM = MB \).

3. **c)** To show that \( DI \perp AC \) at point \( P \):
- You can use the definitions of perpendicular lines and angles, as well as the properties of the square to validate that \( DI \) forms a right angle with \( AC \).

When working through these exercises, be sure to include proper diagrams where necessary and note the use of theorems related to triangle congruence and properties of squares.