Parallelograms: opposite sides and angles equal
The first quadrilateral theorem, built from one diagonal and the congruence tools this course has already earned.
Every theorem so far has been about triangles. This one is about a four-sided shape — but the proof, once it starts, is a triangle proof after all.
A parallelogram, A, B, C and D. A, B and D are draggable; C always follows, to keep the shape a genuine parallelogram.
A parallelogram, A, B, C and D, where A, B and D are draggable and C always follows to keep the shape a genuine parallelogram. AB always matches DC, AD always matches BC, and the angles at opposite corners always match each other too, however the shape is dragged. This needs JavaScript switched on.
Nothing here is forced to look that way by the picture: each of those four numbers is measured independently, off the actual figure, and they still always pair up. A quadrilateral with both pairs of opposite sides parallel is a parallelogram. This lesson’s claim: in any parallelogram, opposite sides are equal, and opposite angles are equal.
One diagonal, one pair of congruent triangles
Draw the diagonal AC. It crosses both pairs of parallel sides at once, as a transversal: AB is parallel to DC, so angle BAC equals angle DCA (alternate angles, agreed two phases ago); AD is parallel to BC, so angle BCA equals angle DAC (alternate angles again, the same fact, the other pair of sides). AC equals itself, shared by both triangles. Two angles and the side between them: ASA, so triangle ABC is congruent to triangle CDA.
What that congruence hands over
Congruent triangles hand over every corresponding part. AB corresponds to CD, so they are equal; BC corresponds to DA, so they are equal too. The angle at B corresponds to the angle at D, so those match directly. The angle at A is angle BAC plus angle CAD; the angle at C is angle DCA plus angle ACB. The congruence already showed angle BAC equals angle DCA and angle CAD equals angle ACB, so adding each pair gives angle A equals angle C as well.
A quadrilateral is not a new kind of problem here — it is two triangles that happen to share an edge, and this course has already built everything two triangles need. One diagonal turns an unfamiliar four-sided claim into a familiar three-sided one, the same move the angle sum of any polygon used for a different quadrilateral fact. The tools do not change; what changes is noticing where to draw the one line that makes them apply.
The next lesson uses the other diagonal fact about a parallelogram — that the two diagonals bisect each other — proved the same way, with a different pair of congruent triangles built from both diagonals crossing at one point. After that, rectangles, rhombi, squares, kites and trapezia all turn out to be parallelograms (or near misses) with one extra condition added.
A parallelogram has one side of 8 units and an adjacent side of 5 units. What are the other two sides?
8 units and 5 units — the side opposite the 8-unit side must also be 8, and the side opposite the 5-unit side must also be 5, by this lesson’s claim.
A parallelogram has one angle of 70 degrees. What is the angle opposite it, and what can be said about the two angles next to it?
The opposite angle is also 70 degrees. The two adjacent angles are not covered by this lesson’s claim directly, but each pairs with the 70-degree angle as co-interior angles on a transversal crossing the parallelogram’s parallel sides, so each of them is 110 degrees.
Why does this proof use only one diagonal, rather than both?
Because one diagonal is already enough to build the ASA congruence the whole argument rests on. Drawing the second diagonal too would not be wrong, but it would not be doing anything this proof needs — it has its own job, in the next lesson.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.