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To prove that the diagonals of a parallelogram bisect each other, let's consider a parallelogram [tex]\(ABCD\)[/tex] with diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]. We need to show that these diagonals intersect at their midpoints.
### Step-by-Step Proof
1. Label the Intersection Point:
Let the diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] intersect at point [tex]\(O\)[/tex].
2. Coordinate Assignment:
Without loss of generality, let's use a coordinate system for easier calculation. Assign coordinates to the vertices of the parallelogram as follows:
- [tex]\(A(x_1, y_1)\)[/tex]
- [tex]\(B(x_2, y_2)\)[/tex]
- [tex]\(C(x_3, y_3)\)[/tex]
- [tex]\(D(x_4, y_4)\)[/tex]
3. Properties of a Parallelogram:
By the definition of a parallelogram, the opposite sides are equal and parallel:
- [tex]\(AB \parallel CD\)[/tex] and [tex]\(AB = CD\)[/tex]
- [tex]\(BC \parallel AD\)[/tex] and [tex]\(BC = AD\)[/tex]
4. Expressions for Midpoints:
To prove diagonals bisect each other, we need to show that the point [tex]\(O\)[/tex] (the intersection of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]) is the midpoint of both diagonals.
First, let's find the midpoint of diagonal [tex]\(AC\)[/tex]:
- Midpoint of [tex]\(AC\)[/tex] is [tex]\(\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)\)[/tex]
Next, we find the midpoint of diagonal [tex]\(BD\)[/tex]:
- Midpoint of [tex]\(BD\)[/tex] is [tex]\(\left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)\)[/tex]
5. Vector Addition Approach:
Considering vector properties in a parallelogram, we can say:
- [tex]\(\overrightarrow{AB}\)[/tex] should be equal to [tex]\(\overrightarrow{CD}\)[/tex]
- [tex]\(\overrightarrow{AD}\)[/tex] should be equal to [tex]\(\overrightarrow{BC}\)[/tex]
Since [tex]\(AB = CD\)[/tex] and [tex]\(AD = BC\)[/tex], the point [tex]\(O\)[/tex] divides each diagonal into two equal parts.
6. Equality of Midpoints:
Given the properties of the parallelogram, the coordinates of point [tex]\(O\)[/tex] must satisfy both midpoint conditions:
- The coordinates of [tex]\(O\)[/tex] as midpoint of [tex]\(AC\)[/tex] is [tex]\(\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)\)[/tex]
- The coordinates of [tex]\(O\)[/tex] as midpoint of [tex]\(BD\)[/tex] is [tex]\(\left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)\)[/tex]
Since both midpoints must be the same point [tex]\(O\)[/tex], we have:
[tex]\[ \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right) \][/tex]
7. Conclusion:
Hence, [tex]\(O\)[/tex] is equidistant from endpoints of both diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]. Thus, the diagonals of the parallelogram bisect each other, as desired.
### Step-by-Step Proof
1. Label the Intersection Point:
Let the diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] intersect at point [tex]\(O\)[/tex].
2. Coordinate Assignment:
Without loss of generality, let's use a coordinate system for easier calculation. Assign coordinates to the vertices of the parallelogram as follows:
- [tex]\(A(x_1, y_1)\)[/tex]
- [tex]\(B(x_2, y_2)\)[/tex]
- [tex]\(C(x_3, y_3)\)[/tex]
- [tex]\(D(x_4, y_4)\)[/tex]
3. Properties of a Parallelogram:
By the definition of a parallelogram, the opposite sides are equal and parallel:
- [tex]\(AB \parallel CD\)[/tex] and [tex]\(AB = CD\)[/tex]
- [tex]\(BC \parallel AD\)[/tex] and [tex]\(BC = AD\)[/tex]
4. Expressions for Midpoints:
To prove diagonals bisect each other, we need to show that the point [tex]\(O\)[/tex] (the intersection of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]) is the midpoint of both diagonals.
First, let's find the midpoint of diagonal [tex]\(AC\)[/tex]:
- Midpoint of [tex]\(AC\)[/tex] is [tex]\(\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)\)[/tex]
Next, we find the midpoint of diagonal [tex]\(BD\)[/tex]:
- Midpoint of [tex]\(BD\)[/tex] is [tex]\(\left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)\)[/tex]
5. Vector Addition Approach:
Considering vector properties in a parallelogram, we can say:
- [tex]\(\overrightarrow{AB}\)[/tex] should be equal to [tex]\(\overrightarrow{CD}\)[/tex]
- [tex]\(\overrightarrow{AD}\)[/tex] should be equal to [tex]\(\overrightarrow{BC}\)[/tex]
Since [tex]\(AB = CD\)[/tex] and [tex]\(AD = BC\)[/tex], the point [tex]\(O\)[/tex] divides each diagonal into two equal parts.
6. Equality of Midpoints:
Given the properties of the parallelogram, the coordinates of point [tex]\(O\)[/tex] must satisfy both midpoint conditions:
- The coordinates of [tex]\(O\)[/tex] as midpoint of [tex]\(AC\)[/tex] is [tex]\(\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)\)[/tex]
- The coordinates of [tex]\(O\)[/tex] as midpoint of [tex]\(BD\)[/tex] is [tex]\(\left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)\)[/tex]
Since both midpoints must be the same point [tex]\(O\)[/tex], we have:
[tex]\[ \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right) \][/tex]
7. Conclusion:
Hence, [tex]\(O\)[/tex] is equidistant from endpoints of both diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]. Thus, the diagonals of the parallelogram bisect each other, as desired.
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