AMC 10 · 2004 · #24
Grade 8 geometry-2dIn triangle ABC we have AB=7, AC=8, BC=9. Point D is on the circumscribed circle of the triangle so that AD bisects angle BAC. What is the value of CDAD?
Pick an answer.
AMC 10 2004 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: Triangle $ABC$ has side lengths $AB=7$, $AC=8$, $BC=9$. A circle is drawn through all three vertices. The bisector of angle $BAC$ is extended until it hits that circle again at a point $D$. Find the value of $\dfrac{AD}{CD}$.
Givens: $AB = 7$, $AC = 8$, $BC = 9$.; $A$, $B$, $C$ all lie on one circle (the circumscribed circle).; $D$ is on that same circle, and line $AD$ splits angle $BAC$ into two equal angles.; Answer choices: (A) $\dfrac{9}{8}$, (B) $\dfrac{5}{3}$, (C) $2$, (D) $\dfrac{17}{7}$, (E) $\dfrac{5}{2}$.
Unknowns: The ratio $\dfrac{AD}{CD}$.
Understand
Restated: Triangle $ABC$ has side lengths $AB=7$, $AC=8$, $BC=9$. A circle is drawn through all three vertices. The bisector of angle $BAC$ is extended until it hits that circle again at a point $D$. Find the value of $\dfrac{AD}{CD}$.
Givens: $AB = 7$, $AC = 8$, $BC = 9$.; $A$, $B$, $C$ all lie on one circle (the circumscribed circle).; $D$ is on that same circle, and line $AD$ splits angle $BAC$ into two equal angles.; Answer choices: (A) $\dfrac{9}{8}$, (B) $\dfrac{5}{3}$, (C) $2$, (D) $\dfrac{17}{7}$, (E) $\dfrac{5}{2}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
A picture is what unlocks this problem, so Tool #1 (Draw a Diagram) leads. Drawing the circle with the four points $A$, $B$, $D$, $C$ on it shows the single fact everything hinges on: because $AD$ splits angle $A$ into two equal halves, the two arcs $BD$ and $CD$ are equal, so the chords $BD$ and $CD$ are equal. That turns two unknown lengths into one. Tool #4 (Introduce a Variable) names that shared length $x$, and Tool #7 (Identify Subproblems) supplies the one clean relation tying the four sides and the diagonal $AD$ together (Ptolemy's relation for a cyclic quadrilateral). The variable $x$ cancels at the end, which is exactly why the answer is a plain number.
Execute — Answer: B
8.G.A.5 Step 1 Draw the circle and read off equal chords
- Draw the circle through $A$, $B$, $C$ and mark $D$ where the bisector of angle $A$ crosses the circle again.
- Going around the circle the order is $A$, $B$, $D$, $C$.
- Because $AD$ bisects angle $BAC$, the angle $\angle BAD$ equals the angle $\angle CAD$.
- Those two equal angles sit at $A$ and open onto the arcs $BD$ and $CD$, so those two arcs are equal in size.
- Equal arcs are cut off by equal chords, so $BD = CD$.
- The two segments from $D$ to the far ends of side $BC$ are the same length.
💡 Cutting the angle in half aims the bisector at the exact middle of the far arc, so it lands the same distance from $B$ and from $C$.
6.EE.B.6 Step 2 Name the shared length
- Let $CD = x$.
- From Step 1, $BD = x$ as well, so both chords out of $D$ are $x$.
- The length we actually want, $AD$, is still unknown, so leave it as $AD$.
- Now the four numbers on the quadrilateral $ABDC$ are known or named: $AB = 7$, $AC = 8$, $BD = x$, $CD = x$, and the two diagonals are $AD$ (unknown) and $BC = 9$.
💡 One letter now stands for both mystery chords, so there is really only one new number to deal with.
8.EE.C.7 Step 3 Tie the sides together with Ptolemy
- For four points on a circle, Ptolemy's relation says the product of the two diagonals equals the sum of the products of the two pairs of opposite sides.
- In quadrilateral $ABDC$ the diagonals are $AD$ and $BC$, and the opposite-side pairs are $(AB, CD)$ and $(AC, BD)$.
- So $AD \cdot BC = AB \cdot CD + AC \cdot BD$.
- Plug in the numbers: $AD \cdot 9 = 7 \cdot x + 8 \cdot x = 15x$.
- This is a plain linear equation in $AD$; dividing both sides by $9$ gives $AD = \dfrac{15x}{9} = \dfrac{5x}{3}$.
💡 One circle rule links every side to the diagonal at once, so a single equation pins $AD$ down in terms of $x$.
7.RP.A.2 Step 4 Form the ratio and watch x cancel
- The question asks for $\dfrac{AD}{CD}$.
- We have $AD = \dfrac{5x}{3}$ and $CD = x$, so $\dfrac{AD}{CD} = \dfrac{5x/3}{x}$.
- The $x$ divides out, leaving $\dfrac{5}{3}$.
- The unknown size of the triangle never mattered, exactly as expected for a ratio.
- The answer is (B).
💡 Both lengths grew from the same $x$, so their ratio is fixed no matter how large the circle is.
8.G.A.5 Draw the circle through $A$, $B$, $C$ and mark $D$ where the bisector of angle $ 6.EE.B.6 Let $CD = x$. From Step 1, $BD = x$ as well, so both chords out of $D$ are $x$. 8.EE.C.7 For four points on a circle, Ptolemy's relation says the product of the two diag 7.RP.A.2 The question asks for $\dfrac{AD}{CD}$. We have $AD = \dfrac{5x}{3}$ and $CD = x Review
Reasonableness: The answer $\dfrac{5}{3}$ is a clean number with no leftover $x$, which is the right shape for a ratio of two lengths on the same figure. A quick size check backs it up: computing the real lengths gives $\cos A = \dfrac{7^2+8^2-9^2}{2\cdot 7\cdot 8} = \dfrac{2}{7}$, from which $CD \approx 5.61$ and $AD \approx 9.35$, and $9.35 / 5.61 = 1.667 = \dfrac{5}{3}$. It also passes a symmetry sniff test: if the triangle were isosceles with $AB = AC$, the bisector would be a line of symmetry, $D$ would sit opposite the midpoint of $BC$, and Ptolemy would give $AD = \dfrac{2\,AB}{BC}\,CD$ — the same structure with $7+8$ replaced by $2\,AB$.
Alternative: Skip Ptolemy and use similar triangles. Let $E$ be the point where $AD$ crosses side $BC$. Angles $\angle ADC$ and $\angle ABE$ subtend the same arc $AC$, so they are equal, and $\angle DAC = \angle BAE$ because $AD$ bisects angle $A$; that makes triangle $ADC \sim$ triangle $ABE$, giving $\dfrac{AD}{CD} = \dfrac{AB}{BE}$. The Angle Bisector Theorem splits $BC=9$ in the ratio $AB:AC = 7:8$, so $BE = \dfrac{7}{15}\cdot 9 = \dfrac{21}{5}$, and $\dfrac{AD}{CD} = \dfrac{7}{21/5} = \dfrac{5}{3}$. Both roads land on $\dfrac{5}{3}$, but Ptolemy gets there in one equation.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Arguing that equal bisected angles cut off equal arcs, so the chords $BD$ and $CD$ are equal.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the shared chord length $CD = BD = x$ so the two mystery lengths become one variable.)8.EE.C.7Solve linear equations in one variable (Solving $9\,AD = 15x$ for $AD = \dfrac{5x}{3}$ from Ptolemy's relation.)7.RP.A.2Recognize and represent proportional relationships between quantities (Forming the ratio $\dfrac{AD}{CD} = \dfrac{5x/3}{x}$ and cancelling $x$ to get $\dfrac{5}{3}$.)
⭐ Cutting angle $A$ in half aims the line at the middle of the far arc, so $BD = CD$; then one circle rule (Ptolemy) turns all the sides into $AD = \tfrac{5}{3}CD$, and the ratio is $\tfrac{5}{3}$.
⭐ Cutting angle $A$ in half aims the line at the middle of the far arc, so $BD = CD$; then one circle rule (Ptolemy) turns all the sides into $AD = \tfrac{5}{3}CD$, and the ratio is $\tfrac{5}{3}$.
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