Competition · AMC preparation · step 4 of 4

AMC 10 · 2022A · #13

Grade 8 geometry-2d
similar-trianglesisosceles-triangleratio-proportion identify-subproblemseasier-related-problem ↑ Prerequisites: similar-triangles
📏 Medium solution 💡 2 insights
Problem
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.

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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 ∥ BC make triangles ADY and CBY similar, with ratio AY:CY = 2:1, so AD = 2 · BC = 10. Tool #9 (Easier Related Problem) — picking concrete values like AB = 2, AC = 3 would make the picture even more concrete if needed.

1STEP 1

Apply the angle bisector theorem

Angle Bisector Theorem: AP splits BC so BP:PC = AB:AC, hence AB:AC = 2:3 — write AB = 2k, AC = 3k.

BP/PC = AB/AC → 2/3 = AB/AC → AB = 2k, AC = 3k
2STEP 2

Add the auxiliary point

Extend BD to meet AC at Y; AP bisects ∠A and is ⊥ to BY, so ASA gives △AXB ≅ △AXY, hence AY = AB = 2k (Y is B mirrored over AP).

△ AXB ≅ △ AXY → AY = AB = 2k
3STEP 3

Find the remaining piece

Since AY = 2k is less than AC = 3k, Y lies between A and C, so YC = AC − AY = k.

YC = AC - AY = 3k - 2k = k
4STEP 4

Spot the similar triangles

With AD ∥ BC, alternate interior angles and vertical angles at Y give AA similarity: △ADY ∼ △CBY.

AD ∥ BC → △ ADY ∼ △ CBY
5STEP 5

Use the similarity ratio

Corresponding sides give AD:CB = AY:CY = 2:1; with CB = 2 + 3 = 5, AD = 2 × 5 = 10.

AD/CB = AY/CY = 2k/k = 2 → AD = 2 × 5 = 10
6STEP 6

Match against the choices

AD = 10 matches choice (C); the others 8, 9, 11, 12 need ratios the 2:3 split can't produce.

AD = 10 → (C)
Answer
10
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 · BC = 8. With the original BC = 5 it scales to AD = 2 · 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.
💡Key takeaway

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 ∥ BC, similar triangles give AD : BC = AY : YC = 2 : 1, so AD = 2 · 5 = 10.

  • Apply the angle bisector theorem
  • Add the auxiliary point
  • Find the remaining piece
  • Spot the similar triangles
  • Use the similarity ratio
  • Match against the choices

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