The diagonals of a parallelogram bisect each other
The second parallelogram theorem, from the same figure as the last lesson and the same ASA move.
The last lesson drew one diagonal of a parallelogram. This one draws both, and asks about the point where they cross.
The same parallelogram as before, A, B, C and D, with both diagonals drawn and the point M where they meet. A, B and D are draggable; C and M always follow.
A parallelogram, A, B, C and D, with both diagonals drawn and the point M where they cross. A, B and D are draggable; C and M always follow to keep the shape a genuine parallelogram. AM always matches MC, and BM always matches MD, however the shape is dragged. This needs JavaScript switched on.
Both diagonals get bisected by the same single point — a stronger claim than either diagonal being bisected on its own. The point where the two diagonals of a parallelogram cross bisects each of them: it is the midpoint of AC, and it is also the midpoint of BD.
Why one triangle pair proves both midpoints at once
Draw both diagonals, AC and BD, crossing at M. AB is parallel to DC (opposite sides of the parallelogram), so diagonal BD, crossing both, makes equal alternate angles: angle ABM equals angle CDM. The other diagonal, AC, crossing the same pair of parallel sides, gives another equal pair: angle BAM equals angle DCM. AB equals DC — the last lesson’s own claim, about the same two sides of the same figure. Two angles and the side between them: ASA, so triangle ABM is congruent to triangle CDM.
Congruent triangles hand over every corresponding part. AM corresponds to CM, so they are equal — M is the midpoint of AC. BM corresponds to DM, so they are equal too — M is the midpoint of BD as well, from the same one triangle pair.
Once ABCD is known to be a parallelogram, most of what follows about it keeps coming from the same recipe: cut it with one line, match a pair of triangles, and read off whatever congruence hands over.
The next lesson turns this the other way round: rectangles, rhombi, squares, kites and trapezia are all parallelograms (or near misses) with one extra condition added — and some of those extra conditions are stated in terms of the diagonals this lesson just described.
A parallelogram’s diagonals cross at M. AM is 6 units. What is MC?
6 units — M is the midpoint of AC, so MC equals AM.
In the same parallelogram, BD is 14 units long. What is BM?
7 units — M is the midpoint of BD too, so BM is half of the whole diagonal.
Why does this lesson only need one pair of congruent triangles, not two?
Because the one pair, ABM and CDM, already hands over both midpoint claims at once: AM=CM comes from one pair of corresponding sides, and BM=DM comes from the other pair, in the same congruence. A second triangle pair (say ADM and CBM) would prove the same two facts again, not new ones.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.