如何严谨证明平行四边形的对角线永不平行？（限欧氏综合几何方法）

Let’s break this down using classic Euclidean synthetic geometry—no coordinate math, just definitions, axioms, and basic theorems we already know about parallelograms and parallel lines.

First, let’s set the stage with key definitions and theorems we’ll lean on:

• A parallelogram is a non-degenerate quadrilateral (four distinct points, no three collinear) where both pairs of opposite sides are parallel.
• Parallel lines (in Euclidean geometry) are lines that lie in the same plane and never intersect, no matter how far they’re extended.
• Theorem: In any non-degenerate quadrilateral, the diagonals (lines connecting opposite vertices) intersect each other inside the quadrilateral.

We’ll use proof by contradiction here—it’s perfect for "never" statements because we can assume the opposite is true, then show that leads to an impossible result.

Step-by-Step Proof: #

1. Let’s name our parallelogram (ABCD), with (AB \parallel CD) and (AD \parallel BC) (this is the definition of a parallelogram). Let its diagonals be (AC) (connecting (A) to (C)) and (BD) (connecting (B) to (D)).

2. Assume the opposite of our claim: Suppose (AC \parallel BD).

3. Now, recall the definition of parallel lines: they never intersect. But wait—we know from the theorem above that in any non-degenerate quadrilateral (which our parallelogram is), the diagonals must intersect each other inside the figure. That’s a direct contradiction right there!

If you want to avoid relying on the "diagonals intersect" theorem, here’s a more granular contradiction built from first principles:

• Since (AB \parallel CD) (by parallelogram definition), consider transversal (AC) cutting these two lines. Alternate interior angles (\angle BAC) and (\angle DCA) are congruent.
• If (AC \parallel BD), then transversal (CD) cutting these parallel lines would make alternate interior angles (\angle ACD) and (\angle BDC) congruent.
• Combining these, (\angle BAC = \angle BDC). But this would imply point (C) lies on line (BD)—which can’t be true, because (ABCD) is a quadrilateral with distinct vertices (non-degenerate). If (C) were on (BD), (ABCD) would collapse into a triangle or line segment, violating the definition of a parallelogram.

Either way, our initial assumption that (AC \parallel BD) leads to an impossible contradiction. Therefore, the diagonals of a parallelogram can never be parallel.

内容的提问来源于stack exchange，提问作者Viktor K.