Proving Quadrilateral Types

To classify a quadrilateral, work in two stages. First establish it is a parallelogram, then add whichever single property identifies the subtype.

On the coordinate plane, three tools do all the work:

• Distance formula → is a side or diagonal congruent?
• Slope → are sides parallel? Are diagonals perpendicular?
• Midpoint formula → do the diagonals bisect each other?

An efficient route for a coordinate proof:
1. Midpoints of both diagonals match → parallelogram.
2. Diagonal lengths equal → rectangle.
3. Diagonal slopes multiply to −1 → rhombus.
4. Both → square.

Compute only what you need. Finding all four side lengths and all four slopes when two midpoints would settle it wastes time and creates chances for arithmetic slips.

Classification proofs are exercises in choosing an efficient sufficient condition. Any of the parallelogram tests establishes the base type; the diagonal characterisations then distinguish rectangle, rhombus and square with a single additional computation each.

In coordinate proofs the three formulas map onto the three questions: distance answers congruence, slope answers parallelism and perpendicularity, and midpoint answers bisection. The midpoint test is usually the cheapest parallelogram proof, requiring two midpoint computations rather than four distances or four slopes.

For a general placement proof — showing a result holds for *every* figure of a type, not one example — variables are used for coordinates, and the figure should be positioned to simplify the algebra: one vertex at the origin, one side along the x-axis, and coordinates like (2a, 0) and (2b, 2c) chosen so midpoints avoid fractions.

In synthetic proofs the analogous efficiency question is which pair of triangles to prove congruent, since a single congruence plus CPCTC usually delivers the needed property.