Final answer:
In the parallelogram QRST, diagonals QS and TR bisect each other. The proof uses congruent triangles formed by the diagonals, which result from the parallelogram's properties where opposite sides are equal. Congruency by SSS criterion leads to the conclusion that each diagonal bisects the other at the center.
Step-by-step explanation:
To demonstrate that the diagonals QS and TR bisect each other in the parallelogram QRST, it is crucial to recognize the properties of a parallelogram. In a parallelogram, opposite sides are congruent, meaning QR = ST and QT = RS. Moreover, opposite angles are equal, and the diagonals bisect each other.
Consider triangles QTR and SQT which are formed by the diagonal QT. As QT is common in both triangles, and QR = ST and QS = TR due to the parallelogram's properties, these two triangles are congruent by SSS criterion (Side-Side-Side). As a result of the congruency, the angles formed at point T and point R where the diagonals intersect are equal, implying that TR bisects QS.
Now, by applying the same logic on triangles TSR and QRS created by diagonal SR, we can state that the triangles are congruent by virtue of QR = ST and QT = RS, with SR being the common side. This congruency signifies that angle S is equal to angle R, hence QS bisects TR. Therefore, we have proven that in the parallelogram QRST, the diagonals QS and TR bisect each other.