Look at the picture.
If ABC is any triangle and AD bisects (cuts in half) the angle BAC, then
[tex]\dfrac{AB}{DB}=\dfrac{AC}{DC}[/tex]
1. In our triangle we have the proportion:
[tex]\dfrac{AD}{AC}=\dfrac{BD}{BC}[/tex]
We have
AD = 2.25, AC = 3, BC = 4, BD = x.
Substitute:
[tex]\dfrac{2.25}{3}=\dfrac{x}{4}[/tex]
[tex]\dfrac{x}{4}=0.75[/tex] multiply both sides by 4
[tex]\boxed{x=3}[/tex]
2. In our triangle we have the proportion:
[tex]\dfrac{BD}{BA}=\dfrac{CD}{CA}[/tex]
We have
BD = x + 4, BA = 8, CD = 2x + 1, CA = 12.
Substitute:
[tex]\dfrac{x+4}{8}=\dfrac{2x+1}{12}[/tex] cross multiply
[tex]12(x+4)=8(2x+1)[/tex] use distributive property a(b + c) = ab + ac
[tex]12x+48=16x+8[/tex] subtract 48 from both sides
[tex]12x=16x=-40[/tex] subtract 16x from both sides
[tex]-4x=-40[/tex] divide both sides by (-4)
[tex]\boxed{x=10}[/tex]
3. We have the similar triangles (AAA). Therefore the lengths of the sides are in proportion:
[tex]\dfrac{x}{8}=\dfrac{15}{5}[/tex]
[tex]\dfrac{x}{8}=3[/tex] multiply both sides by 8
[tex]\boxed{x=24\ in}[/tex]