Triangle ADE

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

Answer: 9

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About mvtrinh

Retired high school math teacher.
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