Prove that M is the midpoint of LN, given that LM is a diameter, NP is perpendicular to LP, and R is the midpoint of LP.

LM is a diameter of the circle. The angle subtended by a diameter at any point on the circumference is $90^\circ$.
Therefore, if we draw the chord RM, we have:

(Angle in a semicircle)

It is given that NP is drawn perpendicular to LP.

Since both RM and NP are perpendicular to the same line segment LP, they must be parallel to each other.

The diagram shows markings on the line segment LP, indicating that $LR = RP$.
This means that R is the midpoint of the line segment LP.

Consider $\triangle LPN$.

• R is the midpoint of side LP (from Step 3).
• RM is a line segment drawn from R, parallel to side NP (from Step 2), and it intersects side LN at M. According to the converse of the Midpoint Theorem, if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side. Therefore, M must be the midpoint of the line segment LN.

If M is the midpoint of LN, then the length of LM is equal to the length of MN.

This completes the proof.