AMC 10 · 2009 · #9
Grade 8 geometry-2dSegment BD and AE intersect at C, as shown, AB=BC=CD=CE, and ∠A=25∠B. What is the degree measure of ∠D?
Pick an answer.
AMC 10 2009 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: Two segments, $BD$ and $AE$, cross at $C$, so $B$, $C$, $D$ lie on one straight line and $A$, $C$, $E$ lie on another. Four segments are equal: $AB=BC=CD=CE$. Also $\angle A$ is $\tfrac{5}{2}$ times $\angle B$. Find the degree measure of $\angle D$.
Givens: $BD$ and $AE$ are straight segments meeting at $C$, so $B,C,D$ are collinear and $A,C,E$ are collinear; $AB=BC$, which makes triangle $ABC$ isosceles; $CD=CE$, which makes triangle $CDE$ isosceles; $\angle A = \tfrac{5}{2}\,\angle B$ (both angles are in triangle $ABC$); Answer choices: (A) $52.5$, (B) $55$, (C) $57.7$, (D) $60$, (E) $62.5$
Unknowns: The degree measure of $\angle D$ (that is, $\angle CDE$ in the lower-right triangle)
Understand
Restated: Two segments, $BD$ and $AE$, cross at $C$, so $B$, $C$, $D$ lie on one straight line and $A$, $C$, $E$ lie on another. Four segments are equal: $AB=BC=CD=CE$. Also $\angle A$ is $\tfrac{5}{2}$ times $\angle B$. Find the degree measure of $\angle D$.
Givens: $BD$ and $AE$ are straight segments meeting at $C$, so $B,C,D$ are collinear and $A,C,E$ are collinear; $AB=BC$, which makes triangle $ABC$ isosceles; $CD=CE$, which makes triangle $CDE$ isosceles; $\angle A = \tfrac{5}{2}\,\angle B$ (both angles are in triangle $ABC$); Answer choices: (A) $52.5$, (B) $55$, (C) $57.7$, (D) $60$, (E) $62.5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #13 Convert to Algebra, #3 Eliminate Possibilities
The two angles are tied together by the ratio $\angle A = \tfrac52\angle B$, so Tool #4 (Introduce a Variable) names $\angle B = x$ and turns every other angle into an expression in $x$. Tool #1 (Draw a Diagram) reads the figure to see two isosceles triangles sharing the crossing point $C$, and that the base angles pair off. Tool #13 (Convert to Algebra) writes the triangle-angle-sum fact as one equation and solves it. Tool #3 (Eliminate Possibilities) matches the final degree value against the five choices.
Execute — Answer: A
8.G.A.5 Step 1 Name the angle and use the first isosceles triangle
- Let $\angle B = x$ degrees.
- In triangle $ABC$ the equal sides are $AB=BC$, so the base $CA$ has equal base angles at $A$ and $C$: $\angle BAC = \angle BCA$.
- That is, $\angle A = \angle BCA$.
- The problem also gives $\angle A = \tfrac52\angle B = \tfrac52 x$, so $\angle BCA = \tfrac52 x$ as well.
💡 Equal sides sit across from equal angles, so the two base angles of an isosceles triangle must match.
8.EE.C.7 Step 2 Solve for the angles of triangle $ABC$
- The three angles of triangle $ABC$ add to $180$: $\angle A + \angle B + \angle BCA = 180$.
- Substituting $\tfrac52 x + x + \tfrac52 x = 180$ gives $6x = 180$, so $x = 30$.
- Then $\angle B = 30$ and $\angle A = \angle BCA = \tfrac52(30) = 75$.
💡 One equation from the angle-sum fact pins down the single unknown $x$.
7.G.B.5 Step 3 Carry the angle across the crossing point
- The segments $BD$ and $AE$ cross at $C$, so $B,C,D$ form one straight line and $A,C,E$ form another.
- The angle $\angle BCA$ and the angle $\angle DCE$ sit on opposite sides of the crossing, making them vertical angles, which are always equal.
- Therefore $\angle DCE = \angle BCA = 75$.
💡 When two straight lines cross, the angles directly across from each other are equal.
8.G.A.5 Step 4 Use the second isosceles triangle to find $\angle D$
- In triangle $CDE$ the equal sides are $CD=CE$, so its base angles at $D$ and $E$ are equal: $\angle CDE = \angle CED$.
- Their triangle-angle-sum gives $\angle DCE + \angle CDE + \angle CED = 180$, that is $75 + 2\angle D = 180$, so $2\angle D = 105$ and $\angle D = 52.5$.
- That is choice (A); no other choice equals $52.5$, so the answer is (A).
💡 The second triangle is isosceles too, so once its top angle is known the two equal base angles split the rest evenly.
8.G.A.5 Let $\angle B = x$ degrees. In triangle $ABC$ the equal sides are $AB=BC$, so th 8.EE.C.7 The three angles of triangle $ABC$ add to $180$: $\angle A + \angle B + \angle B 7.G.B.5 The segments $BD$ and $AE$ cross at $C$, so $B,C,D$ form one straight line and $ 8.G.A.5 In triangle $CDE$ the equal sides are $CD=CE$, so its base angles at $D$ and $E$ Review
Reasonableness: Check every angle. Triangle $ABC$: $75 + 30 + 75 = 180$. The vertical angle at $C$ carries $75$ into triangle $CDE$, whose angles are $75 + 52.5 + 52.5 = 180$. Both triangles close, and $\angle D = 52.5$ is a base angle of a fairly 'wide' apex ($75^\circ$), so it should be a bit less than half of $180$ — $52.5$ fits. It lands exactly on choice (A).
Alternative: Skip naming a variable and reason in fixed steps: since $\angle A = \angle BCA$ and $\angle A = \tfrac52\angle B$, the angle sum $2\angle A + \angle B = 180$ becomes $5\angle B + \angle B = 180$, so $\angle B = 30$ and $\angle BCA = 75$. Vertical angles give $\angle DCE = 75$, and the isosceles triangle $CDE$ yields $\angle D = (180-75)/2 = 52.5$.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about the angle sum of triangles (Applying the triangle angle-sum fact and the isosceles base-angle fact in triangles $ABC$ and $CDE$.)8.EE.C.7Solve linear equations in one variable (Solving $\tfrac52 x + x + \tfrac52 x = 180$ to get $x = 30$ and the angles of triangle $ABC$.)7.G.B.5Use facts about vertical angles to solve for an unknown angle in a figure (Concluding $\angle DCE = \angle BCA = 75$ from the vertical angles formed where $BD$ and $AE$ cross.)
⭐ Name one angle, use 'equal sides mean equal base angles' plus the $180^\circ$ triangle rule, then hop the angle across the crossing point with vertical angles to solve the second triangle.
⭐ Name one angle, use 'equal sides mean equal base angles' plus the $180^\circ$ triangle rule, then hop the angle across the crossing point with vertical angles to solve the second triangle.
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