Question

Julia is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. She starts by assigning coordinates as given.

Drag and drop the correct answer into each box to complete the proof.

The coordinates of point C are (__, c).

The coordinates of the midpoint of diagonal AC¯¯¯¯¯ are (__, c/2 ).

The coordinates of the midpoint of diagonal BD¯¯¯¯¯ are ( a+b/2, __).

AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E with coordinates ​ (a+b/2, c/2) ​ .

By the definition of midpoint, AE¯¯¯¯¯≅ __ and BE¯¯¯¯¯≅ __.

Therefore, diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ bisect each other.

Options:

1. a + b
2. a + c
3. b + c
4. a+b/2
5. a−b/2
6. a/2
7. b/2
8. c/2
9. AC¯¯¯¯¯
10. BD¯¯¯¯¯
11. CE¯¯¯¯¯
12. DE¯¯¯¯¯

Answer 1

Explanation:

Answer 2

1.
`C= (a+b, c)` 1 2.
`E=((a+b)/(2),(c)/(2))` 4 3.
`(c)/(2)` 8 4. AC, BD 5. CE, DE 6.

Explanation:

Below each answer the respective explanation, and also a numerical coordinate parallelogram just to make it clearer.

According to the graph

A= (0,0)

B= (a,0)

C= (a+b, c)

D=(b,c)

E=(a+b/2, c/2)

1. The coordinates of point C are
`\overline{AB}=\overline{CD} = \Rightarrow (0-a) \Rightarrow a`

To find the x-coordinate of C: just add a to b

`C= (a+b, c)`

2. The coordinates of the midpoint of diagonal AC are (__, c/2 ).

Midpoints are calculated by:

`M=\left ( (x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2) \right )`

Since A (0,0) and C (a+b,c) Then: (a+b/2, c/2)

`E=((a+b)/(2),(c)/(2))`

3. The coordinates of the midpoint of diagonal BD are (a+b/2, __).

Then
`(c)/(2)` (8)

4. ___ and ___ intersect at point E with coordinates ​ (a+b/2, c/2)

AC and BD . True. This is a consequence of steps 2 and 3.

5. By the definition of midpoint, AE≅ __ and BE≅ __.

`\overline{AE} \cong\overline{CE}\: and\: \overline{BE}\cong \overline{DE}`

6.Therefore, diagonals AC and BD bisect each other.

In this sense, to bisect is to equally divide into two congruent line segments.