Answer:
C
Explanation:
EFGH is a kite, so EF ≅ FG and EH ≅ HG.
The area of the kite consists of two area of triangles EFG and EHG.
1. Area of triangle EFG:
![A_(\trangle EFG)=(1)/(2)\cdot EG\cdot h,](https://img.qammunity.org/2020/formulas/mathematics/middle-school/x7l4xqgvflqks2mff5m5boifpme8kyq5ij.png)
where h is the height drawn from point F to the side EG.
1. Area of triangle EHG:
![A_(\trangle EHG)=(1)/(2)\cdot EG\cdot H,](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pydzkt16gqhhhyv2bfsgz1pvufvo88tivm.png)
where H is the height drawn from point H to the side EG.
3. Note that
![EG \cong \text{rectangle's length}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7v54shtypz1uaoa29rwgtgcd9t6pnwenbx.png)
![h+H\cong \text{rectangle's width}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z1rnih4fgttrebsclb64yw9eve2u0qnpiu.png)
So,
![A_{\text{kite }EFGH}\\ \\=A_(\triangle EFG)+A_(\triangle EHG)\\ \\=(1)/(2)\cdot EG\cdot (h+H)\\ \\=(1)/(2)\cdot \text{rectangle's length}\cdot \text{rectangle's width}\\ \\=(1)/(2)A_{\text{rectangle}}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qjbarft6wb7uo5u5obrj21nfaq3t7vavav.png)
Thus, option C is true (the area of the kite doesn't depend on ratio in which points E and G divide the side)