Answer:
[tex]AG=22[/tex]
Step-by-step explanation:
Follow the next steps:
[tex]\frac{A-B}{A-E} =\frac{B-C}{E-F} =\frac{C-D}{F-G} =\frac{A-C}{A-F} =\frac{B-D}{E-G} =\frac{A-D}{A-G}[/tex]
Let:
[tex]\frac{A-B}{A-E} =\frac{B-C}{E-F}\\ \\\frac{4}{A-E} =\frac{5}{10x}\\ \\Solving\hspace{3}for\hspace{3}A-E\\\\A-E=8x[/tex]
Now:
[tex]\frac{C-D}{F-G} =\frac{A-C}{A-F} \\\\\frac{2}{F-G} =\frac{9}{18x} \\\\Solving\hspace{3}for\hspace{3}F-G\\\\F-G=4x[/tex]
Hence:
[tex]A-G=(A-E)+(E-F)+(F-G)=22x[/tex]
Finally:
[tex]\frac{B-D}{E-G} =\frac{A-D}{A-G}\\\\\frac{A-D}{B-D} =\frac{A-G}{E-G}\\[/tex]
[tex]\frac{11}{7} =\frac{22x}{14x} \\\\\frac{11x^{2} }{7} -\frac{11}{7} =0\\\\[/tex]
Hence:
[tex]x=1\\x=-1[/tex]
Since it would be absurd for [tex]x=-1[/tex], the real solution is [tex]x=1[/tex]
Therefore:
[tex]AG=22[/tex]
