AMC 8 · 2006 · #19
Grade 8 geometry-2d
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Draw a Diagram) is the natural opening: redraw the figure and mark every piece of information — the two equal sides AB = BC, the two midpoint splits at D, and the known length CE = 11. Tool #12 (Use Symmetry) reads the congruence △ ABD ≅ △ ECD off the vertex order: the matching pairs are A ⇔ E, B ⇔ C, D ⇔ D, so AB matches EC. That single matching turns the given CE = 11 into AB = 11, and the isosceles condition plus the midpoint cuts BC exactly in half. No algebra is needed — three sentences of side-chasing finish it.
Redraw △ ABC with AB = BC, place E so D is the midpoint of both BC and AE, and label the one known length CE = 11.
Drawing and labeling each given is the Grade 4 "identify line segments" move that keeps the equal pieces visible.
4.G.A.1Draw A DiagramThe vertex order in △ ABD ≅ △ ECD pairs A–E, B–C, D–D, so side AB matches side EC — giving AB = 11.
Grade 8 congruence: corresponding parts of congruent triangles are equal. The vertex order is the dictionary that tells you which side equals which.
8.G.A.2Draw A Venn DiagramThe triangle is isosceles with AB = BC, and AB = 11, so BC = 11 too.
An isosceles triangle labels two sides as equal — once one side is known, the matching side is known too.
4.G.A.2Draw A DiagramD is the midpoint of BC, so BD is half of BC: BD = = 5.5.
A midpoint is the Grade 5 "half of a length" idea: cut 11 in half to get 5.5.
5.NF.B.4Draw A DiagramCongruent triangles match corner-to-corner in the order they are written. Once AB = EC = 11, the isosceles side BC is also 11, and the midpoint cuts it in half.