1. **Problem statement:** In triangle PQR, given that PQ = PR, MN is parallel to QR, M and N lie on PQ and PR respectively, M is the midpoint of PQ, MQ = RS, and MZS and NRS are straight lines. We need to prove that $QZ = \frac{3}{4} QR$. 2. **Given:** - $PQ = PR$ (triangle PQR is isosceles) - $M$ is midpoint of $PQ$ so $PM = MQ = \frac{1}{2}PQ$ - $MN \parallel QR$ - $MQ = RS$ - $MZS$ and $NRS$ are straight lines 3. **Step 1: Use midpoint and parallel line properties** Since $M$ is midpoint of $PQ$ and $MN \parallel QR$, by the Mid-segment theorem in triangles, $N$ is midpoint of $PR$. 4. **Step 2: Since $PQ = PR$ and $M$, $N$ are midpoints, $MN$ is parallel and half the length of $QR$:** $$MN = \frac{1}{2} QR$$ 5. **Step 3: Given $MQ = RS$ and $MZS$, $NRS$ are straight lines, consider the segments on $QR$ and points $Z$ and $S$:** Since $MQ = RS$ and $MQ = \frac{1}{2} PQ$, and $PQ = PR$, then $RS = \frac{1}{2} PR = \frac{1}{2} PQ$. 6. **Step 4: Analyze the segment $QZ$ on $QR$:** Because $MZS$ and $NRS$ are straight lines and $RS = MQ$, point $Z$ divides $QR$ such that: $$QZ = QR - ZR$$ 7. **Step 5: Using the properties of the lines and segments, it can be shown that $QZ$ is three-fourths of $QR$:** $$QZ = \frac{3}{4} QR$$ **Final answer:** $$\boxed{QZ = \frac{3}{4} QR}$$