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Hint: - Here, we made a quadrilateral whose one diagonal is x axis. And then suppose the coordinate of the corner of the quadrilateral with respect to x axis and y axis. And then go through the bisector formula of coordinates.

Let ABCD be the quadrilateral such that diagonal AC is along x axis suppose the coordinates A, B, C and D be \[(0,0),({x_2},{y_2})({x_1},0)\]and \[({x_3},{y_3})\] respectively.

E and F are the mid points of sides AD and BC respectively, G and H are the midpoint of diagonals AC and BD. And the point of intersection of EF and GH is I.

Coordinates of E are \[\left( {\dfrac{{0 + {x_3}}}{2},\dfrac{{0 + {y_3}}}{2}} \right) = \left( {\dfrac{{{x_3}}}{2},\dfrac{{{y_3}}}{2}} \right)\]

Coordinates of Fare\[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{0 + {y_2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_2}}}{2}} \right)\]

Coordinates of midpoint of EF are \[\left( {\dfrac{{\dfrac{{{x_3}}}{2} + \dfrac{{{x_1} + {x_2}}}{2}}}{2},\dfrac{{\dfrac{{{y_3}}}{2} + \dfrac{{{y_2}}}{2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{4},\dfrac{{{y_2} + {y_3}}}{4}} \right)\]

G and H are the mid points of diagonal AC and BD respectively then

Coordinates of G are\[\left( {\dfrac{{0 + {x_1}}}{2},\dfrac{{0 + 0}}{2}} \right) = \left( {\dfrac{{{x_1}}}{2},0} \right)\]

Coordinates of H are\[\left( {\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2}} \right)\]

Coordinates of midpoint of GH are \[\left( {\dfrac{{\dfrac{{{x_1}}}{2} + \dfrac{{{x_2} + {x_3}}}{2}}}{2},\dfrac{{\dfrac{{{y_2}}}{2} + \dfrac{{{y_2} + {y_3}}}{2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{4},\dfrac{{{y_2} + {y_3}}}{4}} \right)\]

As you can see, midpoints of both EF and GH are the same. So, EF and GH meet and bisect each other.

Hence, proved.

Note:-Whenever we face such types of questions first of all make the diagram by the statement given by the question, then use the section formula to find the midpoint of a line. If the two points have the same value then it must coincide.

Let ABCD be the quadrilateral such that diagonal AC is along x axis suppose the coordinates A, B, C and D be \[(0,0),({x_2},{y_2})({x_1},0)\]and \[({x_3},{y_3})\] respectively.

E and F are the mid points of sides AD and BC respectively, G and H are the midpoint of diagonals AC and BD. And the point of intersection of EF and GH is I.

Coordinates of E are \[\left( {\dfrac{{0 + {x_3}}}{2},\dfrac{{0 + {y_3}}}{2}} \right) = \left( {\dfrac{{{x_3}}}{2},\dfrac{{{y_3}}}{2}} \right)\]

Coordinates of Fare\[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{0 + {y_2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_2}}}{2}} \right)\]

Coordinates of midpoint of EF are \[\left( {\dfrac{{\dfrac{{{x_3}}}{2} + \dfrac{{{x_1} + {x_2}}}{2}}}{2},\dfrac{{\dfrac{{{y_3}}}{2} + \dfrac{{{y_2}}}{2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{4},\dfrac{{{y_2} + {y_3}}}{4}} \right)\]

G and H are the mid points of diagonal AC and BD respectively then

Coordinates of G are\[\left( {\dfrac{{0 + {x_1}}}{2},\dfrac{{0 + 0}}{2}} \right) = \left( {\dfrac{{{x_1}}}{2},0} \right)\]

Coordinates of H are\[\left( {\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2}} \right)\]

Coordinates of midpoint of GH are \[\left( {\dfrac{{\dfrac{{{x_1}}}{2} + \dfrac{{{x_2} + {x_3}}}{2}}}{2},\dfrac{{\dfrac{{{y_2}}}{2} + \dfrac{{{y_2} + {y_3}}}{2}}}{2}} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{4},\dfrac{{{y_2} + {y_3}}}{4}} \right)\]

As you can see, midpoints of both EF and GH are the same. So, EF and GH meet and bisect each other.

Hence, proved.

Note:-Whenever we face such types of questions first of all make the diagram by the statement given by the question, then use the section formula to find the midpoint of a line. If the two points have the same value then it must coincide.

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