In parallelogrammic figures the opposite sides and angles are equal to one another, and a diagonal cuts them in half. * Let $ACDB$ be a parallelogrammic figure, and $BC$ its diagonal. * I say that for parallelogram $ACDB$, the opposite sides and angles are equal to one another, and the diagonal $BC$ cuts it in half.
The opposite sides and the opposite angles of a parallelogram are equal to one another and either diagonal bisects the parallelogram. In particular, the area of the parallelogram \(\boxdot{ABDC}\) is double the area of \(\triangle{ACB}\), (respectively \(\triangle{BCD}\)).
Proofs: 1 Corollaries: 1 2 3 4 5 6
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