AMC 10 · 2009 · #23

Grade 8 geometry-2d

area-trianglessimilar-trianglesratio-proportion identify-subproblems ↑ Prerequisites: area-triangles

📏 Medium solution 💡 3 insights

Problem

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ . Diagonals $AC$ and $BD$ intersect at $E$ , $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$ ?

Pick an answer.

(A)

$\frac {9}{2}$

(B)

$\frac {50}{11}$

(C)

$\frac {21}{4}$

(D)

$\frac {17}{3}$

(E)

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.

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Toolkit + CCSS Solution

Understand

Restated: A convex quadrilateral ABCD has sides AB = 9 and CD = 12. Its diagonals AC and BD cross at a point E, and the whole diagonal AC has length 14. The two triangles AED and BEC (each made by the crossing diagonals) have the same area. Find the length AE, the part of diagonal AC from A to the crossing point.

Givens: ABCD is convex, with diagonals AC and BD meeting at E.; AB = 9 and CD = 12.; AC = 14, so AE + EC = 14.; Triangle AED and triangle BEC have equal areas.

Unknowns: The length AE.

Understand

Givens: ABCD is convex, with diagonals AC and BD meeting at E.; AB = 9 and CD = 12.; AC = 14, so AE + EC = 14.; Triangle AED and triangle BEC have equal areas.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #4 Introduce a Variable

No picture is given, so I draw the quadrilateral with its two diagonals. The crossing point E splits the figure into four triangles, and the equal-area clue is a statement about two of them. On the diagram I can add a shared triangle to both, which turns the area clue into a parallel-sides fact. Parallel sides then hand me a pair of similar triangles, and the side ratio finishes the problem.

Execute — Answer: E

#7 Identify Subproblems 6.G.A.1 Step 1

Turn equal areas into parallel sides

The diagonals split ABCD into four triangles.
Add the shared triangle DEC to each of the two equal triangles: triangle AED plus triangle DEC makes triangle ADC, and triangle BEC plus triangle DEC makes triangle BDC.
Since AED and BEC start equal, the sums ADC and BDC are equal too.
Triangles ADC and BDC sit on the same base DC, so equal area forces equal height.
That means A and B are the same distance from line DC, which happens exactly when AB is parallel to DC.
So ABCD is a trapezoid with AB parallel to DC.

$$[AED]=[BEC]\Rightarrow [AED]+[DEC]=[BEC]+[DEC]\Rightarrow [ADC]=[BDC]\Rightarrow AB\parallel DC$$

💡 Two triangles on the same base with equal area must have the same height, so their far vertices lie on a line parallel to that base.

#1 Draw a Diagram 8.G.A.5 Step 2

Read off the similar triangles

Look at triangles ABE and CDE, formed where the diagonals cross.
Because AB is parallel to DC, diagonal AC is a line cutting the two parallel lines, so angle BAE equals angle DCE (alternate interior angles).
The same reasoning with diagonal BD gives angle ABE equal to angle CDE.
Two equal pairs of angles are enough, so triangle ABE is similar to triangle CDE.

$$AB\parallel DC\Rightarrow \angle BAE=\angle DCE,\ \angle ABE=\angle CDE\Rightarrow \triangle ABE\sim\triangle CDE$$

💡 Parallel sides cut by the diagonals create matching angles, and matching angles mean the triangles are scaled copies.

#4 Introduce a Variable 8.G.A.4 Step 3

Set the similarity ratio

In the similarity, side AB matches side CD, side AE matches side CE, and side BE matches side DE.
So all these pairs share one scale factor.
That factor is AB over CD, which is 9 over 12, or 3 over 4.
Therefore AE over CE is also 3 over 4.

$$\frac{AE}{CE}=\frac{AB}{CD}=\frac{9}{12}=\frac{3}{4}$$

💡 Similar triangles shrink every matching side by the same factor, so one known pair fixes the ratio of the others.

#4 Introduce a Variable 7.RP.A.3 Step 4

Split AC in the ratio 3 to 4

Now AE and CE are the two pieces of diagonal AC, and their lengths are in the ratio 3 to 4.
Write AE as 3t and CE as 4t.
Their sum is the whole diagonal, 14, so 3t plus 4t is 7t, which equals 14, giving t equal to 2.
Then AE is 3 times 2, which is 6.
So the answer is (E).

$$AE=3t,\ CE=4t,\ 3t+4t=14\Rightarrow t=2\Rightarrow AE=3t=6$$

💡 When a whole is split in a known ratio, count the equal parts and share the total among them.

[1] #7 6.G.A.1 The diagonals split ABCD into four triangles. Add the shared triangle DEC to eac

[2] #1 8.G.A.5 Look at triangles ABE and CDE, formed where the diagonals cross. Because AB is p

[3] #4 8.G.A.4 In the similarity, side AB matches side CD, side AE matches side CE, and side BE

[4] #4 7.RP.A.3 Now AE and CE are the two pieces of diagonal AC, and their lengths are in the ra

Review

Reasonableness: If AE = 6 then CE = 8, and 6 + 8 = 14 matches AC, while 6 to 8 reduces to 3 to 4 as required. It also makes sense that AE is the shorter piece: AB (9) is shorter than CD (12), so the crossing point sits nearer the shorter side AB, giving AE < CE, and indeed 6 < 8. The value lands exactly on a listed choice with no rounding, which is what a clean AMC answer should do.

Alternative: Skip the trapezoid picture and use the sine area formula. Triangles AED and BEC each have their angle at E, and those angles are vertical angles, hence equal. Equal areas give (1/2)(AE)(ED)sin = (1/2)(BE)(EC)sin, so AE times ED equals BE times EC, i.e. AE/EC = BE/ED. Triangles AEB and DEC (vertical angles at E, and AE/EC = BE/ED) are then similar with ratio AB/CD = 9/12 = 3/4, so AE/EC = 3/4 and AE = (3/7)(14) = 6, the same result.

CCSS standards used (min grade 8)

6.G.A.1 Find the area of triangles and other figures by composing and decomposing (Adding the shared triangle DEC to each equal triangle to compare areas of triangles ADC and BDC, and using equal base with equal area to force equal height.)
8.G.A.5 Use informal angle arguments, including the angle-angle criterion for similarity of triangles (Getting equal alternate-interior angles from AB parallel to DC and concluding triangle ABE is similar to triangle CDE by angle-angle.)
8.G.A.4 Understand similarity: corresponding sides of similar figures are proportional (Turning the similarity of triangles ABE and CDE into the equal ratio AE/CE = AB/CD = 3/4.)
7.RP.A.3 Use proportional relationships to solve multistep ratio problems (Splitting the length 14 of AC into parts with ratio 3 to 4 to get AE = 6.)

⭐ Equal triangle areas often hide a pair of parallel sides, and parallel sides give similar triangles whose matching sides share one ratio.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Turn equal areas into parallel sides

Read off the similar triangles

Set the similarity ratio

Split AC in the ratio 3 to 4

Review

CCSS standards used (min grade 8)

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