Look at the picture.
If ABC is any triangle and AD bisects (cuts in half) the angle BAC, then
[tex]\dfrac{AB}{BD}=\dfrac{AC}{DC}[/tex]
In our triangle we have the proportion:
[tex]\dfrac{EG}{GD}=\dfrac{EH}{DH}[/tex]
We have:
[tex]EG=99.2\ ft\\\\GD=62\ ft\\\\EH=112\ ft\\\\DH=(x+2)\ ft[/tex]
Substitute:
[tex]\dfrac{99.2}{62}=\dfrac{112}{x+2}\qquad\text{cross multiply}\\\\99.2(x+2)=(62)(112)\\\\99.2(x+2)=6944\qquad\text{divide both sides by 99.2}\\\\x+2=70\qquad\text{subtract 2 from both sides}\\\\\boxed{x=68\ ft}[/tex]
