Use the right triangle to turn the parallelogram into a rectangle. So I'm thinking of a parallelogram that is both a rectangle and a rhombus. 4. This means we are looking for whether or not both pairs of opposite sides of a quadrilateral are congruent. The second angle pair you’d need for ASA consists of angle DHG and angle FJE. It suffices show that $AC$ bisects $BD$. Step-by-step explanation: * Lets revise the cases of congruent - SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ - SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ - ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ Therefore the diagonals of a parallelogram do bisect each other into equal parts. therefore, using the above rule we get
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