AMC 8 · 2006 · #19
Grade 8 geometry-2dProblem
Triangle ABC is an isosceles triangle with AB=BC. Point D is the midpoint of both BC and AE, and CE is 11 units long. Triangle ABD is congruent to triangle ECD. What is the length of BD?
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Isosceles $\triangle ABC$ has $\overline{AB} = \overline{BC}$. Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$. $\overline{CE} = 11$, and $\triangle ABD \cong \triangle ECD$. Find $\overline{BD}$.
Givens: $\triangle ABC$ is isosceles with $\overline{AB} = \overline{BC}$; $D$ is the midpoint of $\overline{BC}$, so $\overline{BD} = \overline{DC}$; $D$ is the midpoint of $\overline{AE}$, so $\overline{AD} = \overline{DE}$; $\overline{CE} = 11$; $\triangle ABD \cong \triangle ECD$; Answer choices: (A) $4$, (B) $4.5$, (C) $5$, (D) $5.5$, (E) $6$
Unknowns: The length of $\overline{BD}$
Understand
Restated: Isosceles $\triangle ABC$ has $\overline{AB} = \overline{BC}$. Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$. $\overline{CE} = 11$, and $\triangle ABD \cong \triangle ECD$. Find $\overline{BD}$.
Givens: $\triangle ABC$ is isosceles with $\overline{AB} = \overline{BC}$; $D$ is the midpoint of $\overline{BC}$, so $\overline{BD} = \overline{DC}$; $D$ is the midpoint of $\overline{AE}$, so $\overline{AD} = \overline{DE}$; $\overline{CE} = 11$; $\triangle ABD \cong \triangle ECD$; Answer choices: (A) $4$, (B) $4.5$, (C) $5$, (D) $5.5$, (E) $6$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #12 Use Symmetry
Tool #1 (Draw a Diagram) is the natural opening: redraw the figure and mark every piece of information — the two equal sides $\overline{AB} = \overline{BC}$, the two midpoint splits at $D$, and the known length $\overline{CE} = 11$. Tool #12 (Use Symmetry) reads the congruence $\triangle ABD \cong \triangle ECD$ off the vertex order: the matching pairs are $A \leftrightarrow E$, $B \leftrightarrow C$, $D \leftrightarrow D$, so $\overline{AB}$ matches $\overline{EC}$. That single matching turns the given $\overline{CE} = 11$ into $\overline{AB} = 11$, and the isosceles condition plus the midpoint cuts $\overline{BC}$ exactly in half. No algebra is needed — three sentences of side-chasing finish it.
Execute — Answer: D
4.G.A.1 Step 1 - Mark up the figure.
- Sketch $\triangle ABC$ with $\overline{AB} = \overline{BC}$, then place $E$ so that $D$ — already the midpoint of $\overline{BC}$ — is also the midpoint of $\overline{AE}$.
- Label the one known length: $\overline{CE} = 11$.
- The picture now shows two triangles meeting at $D$, with $D$ splitting both $\overline{BC}$ and $\overline{AE}$ into equal halves.
💡 Drawing and labeling each given is the Grade 4 "identify line segments" move that keeps the equal pieces visible.
8.G.A.2 Step 2 - Read the congruence in vertex order.
- The notation $\triangle ABD \cong \triangle ECD$ pairs the corners in the order they are written: $A$ with $E$, $B$ with $C$, $D$ with $D$.
- Corresponding sides match in the same order, so $\overline{AB}$ corresponds to $\overline{EC}$.
💡 Grade 8 congruence: corresponding parts of congruent triangles are equal. The vertex order is the dictionary that tells you which side equals which.
4.G.A.2 Step 3 - Carry the length across the isosceles condition.
- The triangle is isosceles with $\overline{AB} = \overline{BC}$, and we just learned $\overline{AB} = 11$, so $\overline{BC} = 11$ as well.
💡 An isosceles triangle labels two sides as equal — once one side is known, the matching side is known too.
5.NF.B.4 Step 4 - Use the midpoint to cut $\overline{BC}$ in half.
- Since $D$ is the midpoint of $\overline{BC}$, the segment $\overline{BD}$ is exactly half of $\overline{BC}$.
💡 A midpoint is the Grade 5 "half of a length" idea: cut $11$ in half to get $5.5$.
4.G.A.1 Mark up the figure. Sketch $\triangle ABC$ with $\overline{AB} = \overline{BC}$, 8.G.A.2 Read the congruence in vertex order. The notation $\triangle ABD \cong \triangle 4.G.A.2 Carry the length across the isosceles condition. The triangle is isosceles with 5.NF.B.4 Use the midpoint to cut $\overline{BC}$ in half. Since $D$ is the midpoint of $\ Review
Reasonableness: Sanity-check the corresponding parts of $\triangle ABD \cong \triangle ECD$. Side $\overline{BD}$ in the first triangle should match $\overline{CD}$ in the second, and indeed both equal $5.5$ because $D$ is the midpoint of $\overline{BC}$. Side $\overline{AD}$ should match $\overline{ED}$, which is exactly what the second midpoint condition gives. And $\overline{AB} = \overline{EC} = 11$ is what we used. All three pairs check out, so the congruence is consistent with the given midpoint conditions, and $\overline{BD} = 5.5$ — answer (D). The other choices ($4$, $4.5$, $5$, $6$) all force $\overline{BC} \ne 11$, contradicting the chain $\overline{AB} = \overline{EC} = 11$.
Alternative: Tool #13 (Convert to Algebra): let $\overline{BD} = x$. Then $\overline{BC} = 2x$ (midpoint) and $\overline{AB} = 2x$ (isosceles). The congruence pairs $\overline{AB}$ with $\overline{EC}$, so $2x = 11$, giving $x = 5.5$. Same answer, but the diagram-and-symmetry path skips the variable.
CCSS standards used (min grade 8)
4.G.A.1Draw and identify points, lines, line segments, rays, angles, and perpendicular lines (Redrawing the figure and labeling the equal segments, midpoints, and the known length $\overline{CE} = 11$ so the congruence has something visible to act on.)4.G.A.2Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size (Reading the isosceles condition $\overline{AB} = \overline{BC}$ as a defining property of the triangle, which transfers the length $11$ from $\overline{AB}$ to $\overline{BC}$.)5.NF.B.4Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction (Halving $\overline{BC} = 11$ to get $\overline{BD} = \tfrac{1}{2} \cdot 11 = 5.5$ using the midpoint.)8.G.A.2Understand congruence of two-dimensional figures through a sequence of rotations, reflections, translations, and dilations (Reading the vertex correspondence $A \leftrightarrow E$, $B \leftrightarrow C$, $D \leftrightarrow D$ from $\triangle ABD \cong \triangle ECD$ to conclude $\overline{AB} = \overline{EC} = 11$.)
⭐ Congruent triangles match corner-to-corner in the order they are written. Once $\overline{AB} = \overline{EC} = 11$, the isosceles side $\overline{BC}$ is also $11$, and the midpoint cuts it in half.
⭐ Congruent triangles match corner-to-corner in the order they are written. Once $\overline{AB} = \overline{EC} = 11$, the isosceles side $\overline{BC}$ is also $11$, and the midpoint cuts it in half.
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