AMC 10 · 2022 · #13

Grade 8 geometry-2d

similar-trianglesisosceles-triangleratio-proportion identify-subproblemseasier-related-problem ↑ Prerequisites: similar-triangles

📏 Medium solution 💡 2 insights

Problem

Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$

$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: In scalene triangle $ABC$, the angle bisector from $A$ hits side $BC$ at $P$, with $BP = 2$ and $PC = 3$. Drop a perpendicular from $B$ to line $AP$ and extend it; that line meets the line through $A$ parallel to $BC$ at $D$. Find $AD$.

Givens: $\triangle ABC$ is scalene (all sides distinct); $P$ is on $BC$ with $AP$ bisecting $\angle BAC$; $BP = 2$, $PC = 3$ (so $BC = 5$); Line through $B$ perpendicular to $AP$ meets the line through $A$ parallel to $BC$ at $D$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Unknowns: Length $AD$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #9 Easier Related Problem, #3 Eliminate Possibilities

Tool #1 (Draw): Sketch $A, B, C$, the bisector $AP$, the perpendicular from $B$, and the parallel through $A$. The picture immediately suggests adding the auxiliary point $Y$ where line $BD$ meets line $AC$ — because $AP$ is both an angle bisector at $A$ AND perpendicular to $BY$ at the meeting point, $AP$ is the perpendicular bisector of $BY$ inside the angle, giving an isoceles triangle $ABY$ with $AY = AB$. Tool #7 (Subproblems) splits the work into three clean pieces: (a) Angle Bisector Theorem turns $BP:PC = 2:3$ into $AB:AC = 2:3$; (b) the reflection trick gives $AY = AB$, so $YC = AC - AY = (3-2)$-th of the side; (c) parallel lines $AD \parallel BC$ make triangles $ADY$ and $CBY$ similar, with ratio $AY:CY = 2:1$, so $AD = 2 \cdot BC = 10$. Tool #9 (Easier Related Problem) — picking concrete values like $AB = 2, AC = 3$ would make the picture even more concrete if needed.

Execute — Answer: C

#7 Identify Subproblems 7.RP.A.2 Step 1

Apply the Angle Bisector Theorem on $AP$.
Because $AP$ bisects $\angle BAC$, the foot $P$ divides $BC$ in the ratio of the two adjacent sides: $BP : PC = AB : AC$.
Plugging in $BP = 2, PC = 3$ gives $AB : AC = 2 : 3$.
Write $AB = 2k$ and $AC = 3k$ for some positive $k$.

$$\dfrac{BP}{PC} = \dfrac{AB}{AC} \;\Rightarrow\; \dfrac{2}{3} = \dfrac{AB}{AC} \;\Rightarrow\; AB = 2k,\; AC = 3k$$

💡 Grade 7 proportional reasoning: the bisector splits the opposite side in the same ratio as the adjacent sides.

#1 Draw a Diagram 8.G.A.2 Step 2

Add the auxiliary point: extend the line through $B$ perpendicular to $AP$ until it crosses line $AC$ at point $Y$.
Call $X$ the intersection of $BY$ with $AP$.
In triangles $AXB$ and $AXY$: $\angle BAX = \angle YAX$ (since $AP$ is the bisector), $AX$ is shared, and $\angle AXB = \angle AXY = 90^\circ$.
By ASA, the two right triangles are congruent, so $AY = AB$.

$$\triangle AXB \cong \triangle AXY \;\Rightarrow\; AY = AB = 2k$$

💡 Grade 8 congruence via a reflection across the bisector: $Y$ is just $B$'s mirror image across line $AP$.

#7 Identify Subproblems 6.EE.A.2 Step 3

$Y$ lies on segment $AC$ between $A$ and $C$ (since $AY = 2k < 3k = AC$).
Subtract to get $YC$.

$$YC = AC - AY = 3k - 2k = k$$

💡 Grade 6 algebra: write the missing piece as the whole minus the known part.

#1 Draw a Diagram 8.G.A.5 Step 4

Now use the parallel line $AD \parallel BC$.
Triangles $ADY$ and $CBY$ share vertex $Y$, and the alternate interior angles created by transversals $AC$ and $BD$ across the parallel pair make $\angle DAY = \angle BCY$ and $\angle ADY = \angle CBY$.
So $\triangle ADY \sim \triangle CBY$ (AA similarity).

$$AD \parallel BC \;\Rightarrow\; \triangle ADY \sim \triangle CBY$$

💡 Grade 8 angle facts: parallel lines + a transversal = equal alternate interior angles, which forces AA similarity.

#7 Identify Subproblems 7.G.A.1 Step 5

Read off the similarity ratio using corresponding sides $AY$ and $CY$ (both touch the shared vertex $Y$).
The ratio is $AY : CY = 2k : k = 2 : 1$.
Therefore $AD : CB = 2 : 1$, and with $CB = BP + PC = 2 + 3 = 5$ we get $AD = 2 \times 5 = 10$.

$$\dfrac{AD}{CB} = \dfrac{AY}{CY} = \dfrac{2k}{k} = 2 \;\Rightarrow\; AD = 2 \times 5 = 10$$

💡 Grade 7 scale-drawing: similar triangles scale every side by the same factor, so $AD = 2 \times BC$.

#3 Eliminate Possibilities 6.EE.B.5 Step 6

Match $10$ to the answer choices: it's (C).
The other choices ($8, 9, 11, 12$) correspond to other ratios that the $2:3$ bisector split forbids.

$$AD = 10 \;\Rightarrow\; \textbf{(C)}$$

💡 Final compare to the multiple-choice list — only one matches.

[1] #7 7.RP.A.2 Apply the Angle Bisector Theorem on $AP$. Because $AP$ bisects $\angle BAC$, the

[2] #1 8.G.A.2 Add the auxiliary point: extend the line through $B$ perpendicular to $AP$ until

[3] #7 6.EE.A.2 $Y$ lies on segment $AC$ between $A$ and $C$ (since $AY = 2k < 3k = AC$). Subtra

[4] #1 8.G.A.5 Now use the parallel line $AD \parallel BC$. Triangles $ADY$ and $CBY$ share ver

[5] #7 7.G.A.1 Read off the similarity ratio using corresponding sides $AY$ and $CY$ (both touc

[6] #3 6.EE.B.5 Match $10$ to the answer choices: it's (C). The other choices ($8, 9, 11, 12$) c

Review

Reasonableness: Pick concrete numbers to sanity-check. Let $AB = 2, AC = 3, BC = 5$ — this is a real (degenerate-flat) triangle, so bump one side slightly, e.g. $AB = 2, AC = 3, BC = 4$. The angle bisector still gives $BP:PC = 2:3$. Reflecting $B$ over $AP$ lands $Y$ on $AC$ with $AY = AB = 2$, so $YC = 1$. The similarity ratio is $AY/YC = 2$, so $AD = 2 \cdot BC = 8$. With the original $BC = 5$ it scales to $AD = 2 \cdot 5 = 10$. Magnitudes are reasonable: $AD$ is twice $BC$, which fits the picture of $D$ on the parallel line, far from $A$, on the opposite side of $A$ from $C$.

Alternative: Tool #6 (Guess and Check) with coordinates: place $B = (0,0)$, $C = (5,0)$, choose $A = (1, 2)$ (anywhere off line $BC$). Compute the angle bisector direction at $A$, drop the perpendicular from $B$ onto that line, then intersect with the horizontal line through $A$ (parallel to $BC$). The resulting $D$ lands $10$ units from $A$ along the parallel. Same answer (C), confirming the synthetic similarity argument.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing $AB = 2k$ and $AC = 3k$, then computing $YC = AC - AY = k$ in symbolic form.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching the computed $AD = 10$ against the five answer choices to select (C).)
7.RP.A.2 Recognize and represent proportional relationships between quantities (Reading the Angle Bisector Theorem as a proportion $BP/PC = AB/AC$ and using $2:3$ to label sides.)
7.G.A.1 Solve problems involving scale drawings of geometric figures (Scaling $BC = 5$ by the similarity factor $2$ from $\triangle ADY \sim \triangle CBY$ to get $AD = 10$.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (Using the reflection across the bisector $AP$ to show $\triangle AXB \cong \triangle AXY$ and conclude $AY = AB$.)
8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Using alternate interior angles formed by $AD \parallel BC$ and transversals $AC, BD$ to establish $\triangle ADY \sim \triangle CBY$.)

⭐ This AMC 10 problem only needs Grade 8 reflection-and-parallel-line facts you already know — $AP$ being both an angle bisector AND perpendicular to $BD$ makes $D$'s line a mirror, so the new point $Y$ on $AC$ satisfies $AY = AB$. Combined with the parallel $AD \parallel BC$, similar triangles give $AD : BC = AY : YC = 2 : 1$, so $AD = 2 \cdot 5 = 10$.