Both pairs equal prove it
Four side lengths alone can still force a name. A triangle lands on another triangle when it fits exactly on top.
- AB is ticked equal to DC. BC is ticked equal to AD.
- Diagonal AC is drawn. Both triangles share it.
- SSS lands triangle ABC on triangle CDA exactly.
- Both pairs of opposite sides are now parallel.
The rule
Claim: ABCD is convex. AB equals DC, and BC equals AD. Then ABCD is a parallelogram.
Triangle ABC has sides AB, BC and AC. Triangle CDA has sides DC, AD and AC. AB matches DC. BC matches AD. AC is a side of both triangles. SSS says triangles with three matching pairs of sides fit exactly. So SSS lands one triangle on the other.
Angle BAC lands on angle DCA, so they are equal. They are an alternate pair, so AB is parallel to DC. Angle BCA lands on angle DAC. They are an alternate pair too, so BC is parallel to AD.
The diagonal is not a fifth measurement. It is a step in the proof.
Hinged sticks are joined so they can swing.
| Sides in order | Lean | Parallelogram? |
|---|---|---|
| 6, 3, 6, 3 | Upright | Yes |
| 6, 3, 6, 3 | Leaning | Yes |
| 6, 3, 6, 3 | Leaning further | Yes |
| 6, 3, 5, 3 | Any lean | Never |
Both pairs equal force the name at every lean. They do not fix one exact shape.
Lean the sticks. Then change one stick.
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