We solve compound problems using the properties of parallelograms, rectangles, rhombuses, and squares.
Example ①
Let E, F, G, and H be the midpoints of the sides of quadrilateral ABCD (E on AB, F on BC, G on CD, and H on DA). Prove that EFGH is a parallelogram.
Solution: With respect to diagonal AC, the midpoint theorem gives EF∥AC and EF=AC/2. With respect to diagonal AC, HG∥AC and HG=AC/2. Therefore, EF∥HG and EF=HG. By condition ⑤ (one pair of opposite sides is parallel and equal), EFGH is a parallelogram.
Let E, F, G, and H be the midpoints of the sides of quadrilateral ABCD (E on AB, F on BC, G on CD, and H on DA). Prove that EFGH is a parallelogram.
Solution: With respect to diagonal AC, the midpoint theorem gives EF∥AC and EF=AC/2. With respect to diagonal AC, HG∥AC and HG=AC/2. Therefore, EF∥HG and EF=HG. By condition ⑤ (one pair of opposite sides is parallel and equal), EFGH is a parallelogram.
Key Point
- The sequence midpoint theorem → condition ⑤ for being a parallelogram appears frequently.
- The midpoints of the sides of a rectangle form a rhombus; the midpoints of the sides of a rhombus form a rectangle.
Practice Problem
- What quadrilateral is formed by joining the midpoints of the sides of a rectangle, and why?
Answer and Explanation
- Answer: A rhombus. The diagonals of a rectangle are equal (AC=BD), so the four sides of the quadrilateral formed by joining the midpoints of its sides are all equal (each side = half a diagonal). Therefore, it is a rhombus.