2006 AMC 8 Problem 19

Attempt Problem 19 of the 2006 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Triangle ABCABC is an isosceles triangle with AB‾=BC‾.\overline{AB}=\overline{BC}. Point DD is the midpoint of both BC‾\overline{BC} and AE‾,\overline{AE}, and CE‾\overline{CE} is 1111 units long. Triangle ABDABD is congruent to triangle ECD.ECD. What is the length of BD‾?\overline{BD}?
44
4.54.5
55
5.55.5
66
Answer: D
Concepts:congruence (geometry)isosceles trianglemidpoint
Small hint:
Use the congruent triangles to relate ECEC and ABAB.
Big hint:
Then use AB=BCAB=BC and the fact that DD is the midpoint of BCBC.
Solution:
By the congruency condition, we know that AB=EC=11.AB = EC = 11.
Also from the isosceles condition, we know that BC=AB=11.BC = AB = 11.
Since DD is the midpoint of BC‾,\overline{BC}, we know that BD=BC÷2=5.5.BD = BC \div 2 = 5.5.
Thus, D is the correct answer.
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