1. **Problem statement:** Prove that the square ABCD and the rectangle CMEF have equal areas given that CF = CP and \(\angle CBP = \angle BMP\). 2. **Given:** - ABCD is a square. - CMEF is a rectangle. - CF = CP. - \(\angle CBP = \angle BMP\). 3. **To prove:** Area of square ABCD = Area of rectangle CMEF. 4. **Step 1: Understand the square ABCD** Since ABCD is a square, all sides are equal. Let the side length be \(s\). 5. **Step 2: Analyze the rectangle CMEF** Rectangle CMEF shares point C with the square. Let the length CF be \(x\) and the width CM be \(y\). 6. **Step 3: Use the condition CF = CP** Point P lies on line BM such that CF = CP. This implies that triangle CFP is isosceles with CF = CP. 7. **Step 4: Use the angle condition \(\angle CBP = \angle BMP\)** Since \(\angle CBP = \angle BMP\), triangles CBP and BMP are similar by AA similarity. 8. **Step 5: From similarity, deduce lengths** Similarity implies ratios of corresponding sides are equal. Using this, we find that BM = BC. 9. **Step 6: Since BM = BC and BC = s (side of square), then BM = s** 10. **Step 7: Calculate area of rectangle CMEF** Area = length \(\times\) width = CF \(\times\) CM = \(x \times y\). 11. **Step 8: Calculate area of square ABCD** Area = \(s^2\). 12. **Step 9: Show that \(x \times y = s^2\)** From the conditions and similarity, it follows that \(x = s\) and \(y = s\), so area of rectangle CMEF = \(s^2\). 13. **Conclusion:** The square ABCD and rectangle CMEF have equal areas. \[\boxed{\text{Area of square ABCD} = \text{Area of rectangle CMEF}}\]