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. [tex]C= (a+b, c)[/tex] 1 2. [tex]E=(\frac{a+b}{2},\frac{c}{2})[/tex] 4 3. [tex]\frac{c}{2}[/tex] 8 4. AC,  BD 5. CE, DE 6. Step-by-step explanation:Below each answer the respective explanation, and also a numerical coordinate parallelogram just to make it clearer.According to the graphA= (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 [tex]\overline{AB}=\overline{CD} = \Rightarrow (0-a) \Rightarrow a[/tex]To find the x-coordinate of C: just add a to b[tex]C= (a+b, c)[/tex]2. The coordinates of the midpoint of diagonal AC are (__, c/2 ).Midpoints are calculated by:[tex]M=\left ( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )[/tex]Since A (0,0) and C (a+b,c) Then: (a+b/2, c/2)[tex]E=(\frac{a+b}{2},\frac{c}{2})[/tex]3. The coordinates of the midpoint of diagonal BD are (a+b/2, __).Then [tex]\frac{c}{2}[/tex] (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≅ __.[tex]\overline{AE} \cong\overline{CE}\: and\: \overline{BE}\cong \overline{DE}[/tex]6.Therefore, diagonals AC and BD bisect each other.In this sense, to bisect is to equally divide into two congruent line segments.