Points A, B, C, and D are all on segment AD, in the order listed. Segment AB is congruent to segment CD. Segment BC equals 12 units in length. Also, an additional point E exists that is not on segment AD with segment BE being congruent to segment CE. Segment BE equals 10 units in length. If the perimeter of triangle ADE is twice the perimeter of perimeter of triangle BCE, find the length of segment AB.
Source: mathcontest.olemiss.edu 1/22/2007

SOLUTION

Since E is equidistant from B and C, E is on the perpendicular bisector $\overline{EM}\textup{ of }\overline{BC}$. Because $\overline{AB}\cong\overline{CD}$, M is also the midpoint of $\overline{AD}$. Since $\overline{EM}$ is the perpendicular bisector of $\overline{AD}$, E is equidistant from A and D and $EA=ED$.

Length of ME
In right triangle $BME$,
$BM^2+ME^2=BE^2$
$6^2+ME^2=10^2$ substitute $6$ for $BM$ and $10$ for $BE$
$36+ME^2=100$
$ME^2=64$ subtract $36$ from both sides
$ME=8$

Perimeter of triangle BCE
$BE+EC+CB=10+10+12=32$

Perimeter of triangle ADE
$AE+ED+DA=AE+AE+\left (AB+BC+CD\right )$ substitute $AE$ for $ED$
$= 2AE+\left (AB+12+AB\right )$ substitute $12$ for $BC$ and $AB$ for $CD$
$= 2AE+2AB+12$

Because the perimeter of triangle ADE is twice the perimeter of triangle BCE,
$2\left (32\right )=2AE+2AB+12$
$32=AE+AB+6$
divide both sides by 2
$26=AE+AB$        (1)

Length of AB
In right triangle $AME$,
$AM^2+ME^2=AE^2$
$\left (AB+BM\right )^2+ME^2=AE^2$
$\left (AB+6\right )^2+8^2=AE^2$
(2) substitute $6$ for $BM$ and $8$ for $ME$
Substitute the value of $AE=26-AB$ from Eq. (1) into Eq. (2),
$\left (AB+6\right )^2+64=\left (26-AB\right )^2$
$64=\left (26-AB\right )^2-\left (AB+6\right )^2$ subtract $\left (AB+6\right )^2$ from both sides
$64=\left (26-AB+AB+6\right )\left (26-AB-AB-6\right )$ apply identity $a^2-b^2=\left (a+b\right )\left (a-b\right )$
$64=\left (32\right )\left (20-2AB\right )$ simplify
$2=20-2AB$ divide both sides by $32$
$2AB=18$
$AB=9$