AMC 10 · 2002 · #23

Grade 10 geometry-2d

perpendicular-bisectorisosceles-trianglesimilar-triangles identify-subproblems ↑ Prerequisites: perpendicular-bisectorisosceles-triangle

📏 Long solution 💡 4 insights

Problem

In triangle ABC, side AC and the perpendicular bisector of BC cross at a point D, and the segment BD bisects ∠ ABC. Given AD = 9 and DC = 7, find the area of triangle ABD.

Pick an answer.

(A)

(B)

(C)

(D)

$14\sqrt5$

(E)

$28\sqrt5$

How to solve

Strategy Identify Subproblems

Area needs a base and a height, and the problem hands over neither. Tool #7 (Identify Subproblems) splits the work honestly into two jobs: first pin down the missing side AB, then get a height. Tool #1 (Draw a Diagram) is where the first job starts, because the two hypotheses are really angle facts in disguise — a perpendicular bisector says two lengths are equal, and an isosceles triangle says two angles are equal. Tool #4 (Introduce a Variable) names the one angle that all the others turn out to be copies of. Tool #16 (Change Focus) is the key turn: instead of hunting inside triangle ABD, look at the big triangle ACB next to it and notice the two share enough angles to be similar.

1STEP 1

Read the perpendicular bisector

D on the perpendicular bisector of BC gives DB = DC = 7, making triangle DBC isosceles.

DB = DC = 7 → ∠ DBC = ∠ DCB

2STEP 2

Name one angle and spread it around

Naming θ = ∠ ACB and using the bisector spreads it: ∠ ABD = ∠ DBC = ∠ ACB = θ.

∠ ACB = ∠ DBC = ∠ ABD = θ, ∠ ABC = 2θ

3STEP 3

Spot the similar triangles

A shared angle at A plus that equality gives triangle ABD similar to triangle ACB.

∠ A = ∠ A, ∠ ABD = ∠ ACB → △ ABD ∼ △ ACB

4STEP 4

Turn the similarity into AB

The similarity gives AB² = AD · AC = 9 · 16, so AB = 12.

AB/AC = AD/AB → AB² = AD · AC = 9 · 16 = 144 → AB = 12

5STEP 5

Find the height onto AC

With sides 12, 9, 7 known, dropping a perpendicular from B gives height h = 28√(5)/9.

x²+h²=144, (x-9)²+h²=49 → x=88/9, h=28√(5)/9

6STEP 6

Compute the area

So the area is 1/2 · 9 · h = 14√(5), choice (D).

[ABD] = 1/2 · 9 · 28√(5)/9 = 14√(5) → (D)

Answer

14√5

Two independent checks agree. First, Heron's formula on the sides 12, 9, 7: the semiperimeter is 14, and √(14 · 2 · 5 · 7) = √(980) = 14√(5), matching. Second, the configuration is internally consistent, which is worth verifying because the problem imposes two separate conditions on one point. The angle bisector theorem gives AB/BC = AD/DC = 9/7, so BC = 28/3; in the isosceles triangle DBC the foot of the perpendicular is the midpoint of BC, so cosθ = (BC/2)/DC = (14/3)/7 = 2/3. Independently, the law of cosines in triangle ABD gives cos∠ ABD = (12²+7²-9²)/(2 · 12 · 7) = 112/168 = 2/3. The same value, so nothing in the data conflicts. Size check: a triangle with sides 12 and 9 has area at most 1/2 · 12 · 9 = 54, which immediately kills choice (E) 28√(5)≈ 62.6; and since ∠ ADB is obtuse the triangle is visibly flatter than a right triangle would be, so a value near 31 rather than near 54 is what to expect.

💡Key takeaway

A perpendicular bisector and an angle bisector each hand you a pair of equal angles — chase them until a small triangle copies a big one, and the missing side falls out.

Read the perpendicular bisector
Name one angle and spread it around
Spot the similar triangles
Turn the similarity into AB
Find the height onto AC
Compute the area