Answer:
See explanation
Step-by-step explanation:
a) To prove that DEFG is a rhombus, it is sufficient to prove that:
- All the sides of the rhombus are congruent: [tex]|DG|\cong |GF| \cong |EF| \cong |DE|[/tex]
- The diagonals are perpendicular
Using the distance formula; [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}[/tex]
[tex]\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}[/tex]
[tex]|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}[/tex]
[tex]\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}[/tex]
[tex]|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}[/tex]
[tex]\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}[/tex]
[tex]|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}[/tex]
[tex]\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}[/tex]
Using the slope formula; [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
The slope of EG is [tex]m_{EG}=\frac{2c-0}{0-0}[/tex]
[tex]\implies m_{EG}=\frac{2c}{0}[/tex]
The slope of EG is undefined hence it is a vertical line.
The slope of DF is [tex]m_{DF}=\frac{c-c}{a+b-(-a-b)}[/tex]
[tex]\implies m_{DF}=\frac{0}{2a+2b)}=0[/tex]
The slope of DF is zero, hence it is a horizontal line.
A horizontal line meets a vertical line at 90 degrees.
Conclusion:
Since [tex]|DG|\cong |GF| \cong |EF| \cong |DE|[/tex] and [tex]DF \perp FG[/tex] , DEFG is a rhombus
b) Using the slope formula:
The slope of DE is [tex]m_{DE}=\frac{2c-c}{0-(-a-b)}[/tex]
[tex]m_{DE}=\frac{c}{a+b)}[/tex]
The slope of FG is [tex]m_{FG}=\frac{c-0}{a+b-0}[/tex]
[tex]\implies m_{FG}=\frac{c}{a+b}[/tex]
