Given:
AE=EB=12.
AC=3x-12 and BC=2x+4.
By perpendicular bisector property, we get
[tex]\frac{AC}{AE}=\frac{BC}{EB}[/tex]Substitute AE=EB=12, AC=3x-12 and BC=2x+4, we get
[tex]\frac{3x-12}{12}=\frac{2x+4}{12}[/tex]Cancel out the common term, we get
[tex]3x-12=2x+4[/tex]Adding 12 to both sides of the equation, we get
[tex]3x-12+12=2x+4+12[/tex][tex]3x=2x+16[/tex]Subtracting 2x from both sides of the equation, we get
[tex]3x-2x=2x+16-2x[/tex][tex]x=16[/tex]Substitute x=16 in AC=3x-12 , we get
[tex]AC=3(16)-12[/tex][tex]AC=36[/tex]We get AC=36 units.
Given that AD=y+16 and DB=3y+22 and AE=EB=12.
By perpendicular bisector property, we get
[tex]\frac{AD}{AE}=\frac{DB}{EB}[/tex]Substitute AD=y+16 and DB=3y+22 and AE=EB=12 in the equation, we get
[tex]\frac{y+16}{12}=\frac{3y+22}{12}[/tex]Cancel out the common terms, we get
[tex]y+16=3y+22[/tex]Subtracting 22 from both sides, we get
[tex]y+16-22=3y+22-22[/tex][tex]y-6=3y[/tex]Subtracting y from both sides of the equation, we get
[tex]y-6-y=3y-y[/tex][tex]-6=2y[/tex]Dividing both sides by 2, we get
[tex]-\frac{6}{2}=\frac{2y}{2}[/tex][tex]y=-3[/tex]Substitute y=-3 in DB=3y+22, we get
[tex]DB=3(-3)+22[/tex][tex]DB=-9+22[/tex][tex]DB=13[/tex]We get DB=13.
Use Pythagorean theorem to find DE.
[tex]DB^2=DE^2+EB^2[/tex]Substitute DB=13 and EB=12 in the equation, we get
[tex]13^2=DE^2+12^2[/tex][tex]DE^2=13^2-12^2[/tex][tex]DE^2=25[/tex]Taking square root on both sides, we get
[tex]DE=\sqrt[]{25}[/tex][tex]DE=\sqrt[]{5^2}[/tex][tex]DE=5[/tex]We get DE=5 units.
Hence the answers are
[tex]\bar{AC}=36\text{units}[/tex][tex]\bar{DB}=13units[/tex][tex]\bar{DE}=5units[/tex]