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AMC 8 · 2006 · #19

Grade 8 geometry-2d
Archetype Arithmetic Expression Decomposition
isosceles-trianglesimilar-triangleslogical-deduction identify-subproblems ↑ Prerequisites: isosceles-triangle
📏 Short solution 💡 3 insights 📊 Diagram

Problem

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 11 units long. Triangle ABDABD is congruent to triangle ECDECD. What is the length of BD‾\overline{BD}?

Pick an answer.

(A)
4
(B)
4.5
(C)
5
(D)
5.5
(E)
6
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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

#1 Draw a Diagram 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.
$$\overline{AB} = \overline{BC}, \quad \overline{BD} = \overline{DC}, \quad \overline{AD} = \overline{DE}, \quad \overline{CE} = 11$$

💡 Drawing and labeling each given is the Grade 4 "identify line segments" move that keeps the equal pieces visible.

#12 Use Symmetry 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 $\overline{EC}$.
$$\triangle ABD \cong \triangle ECD \;\Rightarrow\; \overline{AB} = \overline{EC} = 11$$

💡 Grade 8 congruence: corresponding parts of congruent triangles are equal. The vertex order is the dictionary that tells you which side equals which.

#1 Draw a Diagram 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.
$$\overline{AB} = 11 \text{ and } \overline{AB} = \overline{BC} \;\Rightarrow\; \overline{BC} = 11$$

💡 An isosceles triangle labels two sides as equal — once one side is known, the matching side is known too.

#1 Draw a Diagram 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 $\overline{BC}$.
$$\overline{BD} = \tfrac{1}{2}\,\overline{BC} = \tfrac{11}{2} = 5.5 \;\Rightarrow\; \textbf{(D)}$$

💡 A midpoint is the Grade 5 "half of a length" idea: cut $11$ in half to get $5.5$.

[1] #1 4.G.A.1 Mark up the figure. Sketch $\triangle ABC$ with $\overline{AB} = \overline{BC}$,
[2] #12 8.G.A.2 Read the congruence in vertex order. The notation $\triangle ABD \cong \triangle
[3] #1 4.G.A.2 Carry the length across the isosceles condition. The triangle is isosceles with
[4] #1 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.1 Draw 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.2 Classify 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.4 Apply 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.2 Understand 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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