Given:

• Triangle $ABC$ is a right-angled triangle with a right angle at $C$.
• $M$ is the midpoint of the hypotenuse $AB$.
• A line through $M$ is drawn parallel to $BC$, and this line intersects $AC$ at $D$.

We need to show the following:

1. $D$ is the midpoint of $AC$.
2. $MD \parallel AC$.
3. $CM = \frac{1}{2} AB$.

Proof:

Part (i): Show that $D$ is the midpoint of $AC$

To prove that $D$ is the midpoint of $AC$, we use the Midpoint Theorem and Properties of Parallel Lines.

• The line through $M$ (the midpoint of $AB$) is parallel to $BC$, and it intersects $AC$ at $D$.
• By the Midpoint Theorem, a line drawn parallel to one side of a triangle and passing through the midpoint of another side divides the third side into two equal parts.
• In our case, the line through $M$ is parallel to $BC$ and intersects $AC$ at $D$. Therefore, by the Midpoint Theorem, we conclude that $D$ divides $AC$ into two equal parts, which means that $D$ is the midpoint of $AC$.

Thus, $D$ is the midpoint of $AC$.

---

Part (ii): Show that $MD \parallel AC$

To show that $MD \parallel AC$, we use the fact that parallel lines remain parallel when a transversal intersects them.

• We are given that $MD$ is parallel to $BC$, and by the properties of similar triangles, the parallel line through $M$ will also be parallel to $AC$ because they are both parallel to $BC$.
• Specifically, triangle $ABC$ is divided by the line through $M$ into two smaller triangles, $MDC$ and $ABC$, which are similar by AA similarity (since $\angle DMC = \angle ABC$ and $\angle MDC = \angle ACB$).
• Therefore, since $MD \parallel BC$, and $BC \parallel AC$, it follows that $MD \parallel AC$.

Thus, $MD \parallel AC$.

---

Part (iii): Show that $CM = \frac{1}{2} AB$

We will use a property of right-angled triangles: the median to the hypotenuse of a right-angled triangle is half the length of the hypotenuse.

• In triangle $ABC$, $AB$ is the hypotenuse, and $M$ is the midpoint of $AB$.
• A well-known result is that in a right-angled triangle, the length of the median to the hypotenuse is equal to half the length of the hypotenuse.
• Therefore, we have: $CM = \frac{1}{2} AB$

Thus, $CM = \frac{1}{2} AB$.

---

Conclusion:

• (i) $D$ is the midpoint of $AC$.
• (ii) $MD \parallel AC$.
• (iii) $CM = \frac{1}{2} AB$.

Would you like a more detailed explanation of any of the steps? Here are some related questions to explore further:

1. What is the significance of the midpoint theorem in geometric proofs?
2. How do you prove similarity between triangles?
3. What is the relationship between the angles in similar triangles?
4. How does the concept of medians apply in other types of triangles?
5. Can this proof be generalized for any right-angled triangle?

Tip: When proving properties involving midpoints, always check if the midpoint theorem can be applied, as it often simplifies geometric proofs significantly.