8.3 Properties of Parallelograms (Theorems 8.1 - 8.7)

Given: Quad $ABCD$ where $AC=BD$, $AC \perp BD$, and they bisect each other.

Step 1: Prove it's a Parallelogram. Since diagonals bisect, it is a $||gm$ (Thm 8.7).

Step 2: Prove it's a Rhombus. Since diagonals are perpendicular ($AC \perp BD$), in $\Delta AOD$ and $\Delta AOB$, $OD=OB$, $\angle AOD=\angle AOB=90$, $AO$ common. So $\Delta AOD \cong \Delta AOB$ (SAS). $AD=AB$. Adjacent sides equal $\implies$ Rhombus.

Step 3: Prove it's a Rectangle. In $\Delta ABC$ and $\Delta BAD$: $AB=AB$, $BC=AD$ (opp sides), $AC=BD$ (given). By SSS, $\Delta ABC \cong \Delta BAD$. $\angle ABC = \angle BAD$. Since $AD || BC$, sum is $180$. $2\angle A = 180 \implies \angle A = 90$.

Conclusion: A parallelogram that is a rhombus and a rectangle is a Square.