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ABCD is a rectangle formed by the points A $(-1, -1)$, B $(-1, 6)$, C $(3, 6)$ and D $(3, -1)$. P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.
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Co-ordinates of point P are $(\frac{-1-1}{2}, \frac{-1+6}{2})$ i.e. $(-1, \frac{5}{2})$ Co-ordinates of point Q are $(\frac{-1+3}{2}, \frac{6+6}{2})$ i.e. $(1, 6)$ Co-ordinates of point R are $(\frac{3+3}{2}, \frac{6-1}{2})$ i.e. $(3, \frac{5}{2})$ Co-ordinates of point S are $(\frac{-1+3}{2}, \frac{-1-1}{2})$ i.e. $(1, -1)$ Co-ordinates of mid point of diagonal QS are $(\frac{1+1}{2}, \frac{6-1}{2})$ i.e. $(1, \frac{5}{2})$ Co-ordinates of mid point of diagonal PR are $(\frac{-1+3}{2}, \frac{5/2+5/2}{2})$ i.e. $(1, \frac{5}{2})$ Since coordinates of mid point of QS = coordinates of mid point of PR Therefore, diagonals PR and QS bisect each other.