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In the given problem involving right triangle \( ABC \) with \( M \) as the midpoint of hypotenuse \( AB \), and points \( C \) and \( D \) such that \( DM = CM \), we need to show several properties. Let's address each one systematically:

### (i) \( \triangle AMC \cong \triangle BMD \)

Since \( M \) is the midpoint of \( AB \), we have:
- \( AM = MB \)
- \( \angle AMC = \angle BMD \) (Each is 90° since \( M \) lies on the hypotenuse and \( DM = CM \) makes \( \triangle MCD \) isosceles with \( \angle MCD = \angle MDC \))
- \( CM = DM \) (Given)

Thus, by SAS (Side-Angle-Side) congruence postulate, \( \triangle AMC \cong \triangle BMD \).

### (ii) \( \angle DBC \) is a right angle

Since \( \triangle AMC \cong \triangle BMD \), corresponding parts of congruent triangles are equal, which implies:
- \( \angle DBC = \angle ACB \) (Corresponding angles in congruent triangles)
- \( \angle ACB = 90^\circ \) (As \( \angle ACB \) is the right angle of the right triangle \( ABC \))

Hence, \( \angle DBC \) is a right angle.

### (iii) \( \triangle DBC \cong \triangle ACB \)

For triangles \( \triangle DBC \) and \( \triangle ACB \):
- \( \angle DBC = \angle ACB = 90^\circ \) (From previous proofs)
- \( BC \) is a common side.
- \( BD = AC \) (Since \( \triangle AMC \cong \triangle BMD \), \( BD = AM \) and \( AM = AC \) as \( M \) is the midpoint of \( AB \))

By RHS (Right-angle Hypotenuse-Side) congruence criterion, \( \triangle DBC \cong \triangle ACB \).

### (iv) \( CM = \frac{1}{2} AB \)

Given \( M \) is the midpoint of \( AB \), by definition, \( CM \) (which is \( AM \) or \( MB \)) is exactly half of \( AB \).

Would you like more details on any of these proofs or have any questions about the steps taken? Here are some further questions to explore:

1. Why does the midpoint of the hypotenuse of a right triangle always coincide with the circumcenter?
2. How does the property of triangle congruence help in solving real-world problems?
3. Can you derive the Pythagorean theorem from the given triangle and its properties?
4. What are other significant points within a triangle besides the midpoint of the hypotenuse?
5. How can understanding these geometric principles aid in understanding more complex shapes?

Tip: When proving triangles are congruent, always clearly state which criterion you are using (SSS, SAS, ASA, AAS, or RHS) and ensure you have demonstrated each part of that criterion.

In right triangle ABC, right-angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB.

Solve this geometric problem involving a right triangle ABC, where M is the midpoint of hypotenuse AB, and D is a point such that DM = CM. The problem requires proving triangle congruence and understanding right-angle properties using geometric theorems and congruence criteria. Suitable for students in Grades 9-11.

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