Three more ways in
Three more measurements can force the name too.
- Diagonals AC and BD cross at O.
- OA is ticked equal to OC. OB is ticked equal to OD.
- Angles AOB and COD are equal. They are vertically opposite.
- SAS lands triangle AOB on triangle COD. AB is parallel to DC.
Three more routes
One pair, equal and parallel. Suppose AB equals DC. Suppose AB is parallel to DC too. Then angles BAC and DCA are equal alternate angles. SAS: two sides and the angle between them match. Here AB equals DC, and AC is shared. Angle BAC equals angle DCA. So SAS lands triangle ABC on triangle CDA. "Lands on" means fits exactly on top. Then angles BCA and DAC are equal. So BC is parallel to AD too.
Diagonals that bisect. To bisect is to cut into two equal parts. Suppose OA equals OC, and OB equals OD. Angles AOB and COD are vertically opposite, so they are equal. SAS lands triangle AOB on triangle COD. So AB equals DC. Angles OAB and OCD are equal alternate angles. So AB is parallel to DC. Now the first route finishes the proof.
Opposite angles equal. The four angles total 360°. Suppose angle A equals angle C, and angle B equals angle D. Then A + B + C + D = A + B + A + B = 360°. So A + B = 180°. Angles A and B are co-interior. They lie on one side of AB, between AD and BC. Co-interior angles that add to 180° force AD parallel to BC. Angle D equals angle B, so A + D = 180°. That forces AB parallel to DC.
| Given | What forces the name |
|---|---|
| One pair equal and parallel | SAS, then the other alternate pair |
| Diagonals bisect each other | SAS at O, then the one-pair route |
| Opposite angles equal | The 360° total, then co-interior angles |
Watch out!
One parallel pair is not enough. Parallel sides of 4 and 6 give a trapezium. A trapezium has one parallel pair. The pair must be equal too.
Try the diagonal route. AC and BD cross at O. Turn BD about O. O stays the middle of both diagonals. ABCD stays a parallelogram. Now slide BD off centre. OB and OD differ. ABCD is no longer a parallelogram.
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