AMC 10 Step by Step

Similar & Congruent Triangles

Similarity ratios, congruence, proportional segments, area ratios of similar figures

27
primary-topic problems (2.1% of all)
44
more as a secondary topic
Where it appears
6
P1-10
6
P11-15
11
P16-20
4
P21-25

What you need to know

  • AA similarity. If two angles of one triangle equal two angles of another, the triangles are similar (same shape). Matching sides are all in the same ratio kk, and the areas are in ratio k2k^2. Example: if the sides are 33 times as long, the area is 99 times as big.
  • A line parallel to one side. It cuts the other two sides in the same proportion. Example: in triangle ABCABC with DD on ABAB, EE on ACAC, and DE∥BCDE\parallel BC: if AD=2AD = 2, DB=3DB = 3, and AE=4AE = 4, then ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} gives EC=6EC = 6. Equal angles for AA usually come from parallel lines, vertical angles (opposite angles where two lines cross), or an angle both triangles share.
  • Congruence tests (same shape and same size): SSS (three sides), SAS (two sides and the angle between them), ASA or AAS (two angles and the matching side; a 3030-6060-9090 triangle with short leg 55 is not congruent to one with long leg 55), and HL (hypotenuse and a leg, for right triangles).
  • The altitude to the hypotenuse. In a right triangle with legs aa, bb and hypotenuse cc, the altitude hh to the hypotenuse makes three similar right triangles. If it cuts the hypotenuse into pp (next to leg aa) and qq (next to leg bb), then h2=pqh^2 = pq, a2=pca^2 = pc, and b2=qcb^2 = qc. Example: in the 33-44-55 triangle, p=95p = \frac95 and q=165q = \frac{16}5, so h2=14425h^2 = \frac{144}{25} and h=125h = \frac{12}{5}; also 32=95⋅53^2 = \frac95\cdot5 and 42=165⋅54^2 = \frac{16}{5}\cdot5.

How AMC 10 tests it

  • Problems 8–15: a square or rectangle with a line from a corner through a side, extended to meet another line. The rectangle's parallel sides create the similar pair.
  • A trapezoid with both diagonals: the triangles on the two bases are similar, and the two side triangles have equal area.
  • "DE∥BCDE\parallel BC splits triangle ABCABC into two equal areas; find ADAB\frac{AD}{AB}." Triangle ADEADE has half the area, so k2=12k^2 = \frac12 and k=22k = \frac{\sqrt2}{2}: the 2\sqrt2 comes from area scaling by k2k^2.

Standard approaches

  1. Mark equal angles (parallel lines, vertical angles, shared angles, right angles). Any two give a similar pair.
  2. Write the similarity with matching corners in order, so the ratios can be read off: △ABE∼△FCE\triangle ABE \sim \triangle FCE gives ABFC=BECE\frac{AB}{FC} = \frac{BE}{CE}.
  3. Turn length ratios into area ratios by squaring. Or, for triangles with the same height, the area ratio is the ratio of the bases.

Worked example

Square ABCDABCD has side length 66. Point EE lies on side BCBC with BE=2BE = 2. Line AEAE meets line DCDC, extended past CC, at FF. What is the area of triangle ADFADF?

(A) 3636 (B) 4242 (C) 4848 (D) 5454 (E) 6060

Solution. Step 1. Since AB∥DFAB\parallel DF, the alternate angles ∠BAE\angle BAE and ∠CFE\angle CFE are equal. The vertical angles at EE are equal too. So △ABE∼△FCE\triangle ABE \sim \triangle FCE, with AA, BB, EE matching FF, CC, EE.

Step 2. CE=6−2=4CE = 6 - 2 = 4. Matching sides give
FCAB=CEBE=42=2, \frac{FC}{AB} = \frac{CE}{BE} = \frac{4}{2} = 2,
so FC=2⋅6=12FC = 2\cdot 6 = 12 and DF=6+12=18DF = 6 + 12 = 18.

Step 3. Triangle ADFADF has a right angle at DD, base DF=18DF = 18, and height AD=6AD = 6. Its area is 12⋅18⋅6=54\frac12\cdot 18\cdot 6 = 54.

The answer is (D) 54\boxed{\textbf{(D)}\ 54}.

Pitfalls

  • Pairing sides that don't match, like ABCE\frac{AB}{CE} instead of ABFC\frac{AB}{FC}. Match sides by the angles across from them: ABAB and FCFC are both across from the angle at EE.
  • Forgetting to square the ratio for areas, or to take the square root going from areas back to lengths. Example: sides in ratio 2:32:3 mean areas in ratio 4:94:9.
  • Assuming a trapezoid's diagonals cut each other in half. That is true only for parallelograms.

Traps that recur

  • Reading the ratio 1 : 5 as a division of BE and concluding BC = 7/6; the similarity pairs BC with DC and AC with EC, never BC with CE. (2025 AMC 10A #15)
  • Assuming O is the midpoint of BD (it is not; O splits BD in ratio AB:CD = 2:1), or mixing up which segment OP = 11 measures. (2021 AMC 10A #17)
  • Trying to solve with the side as unknown in two similar triangles and getting a quartic, or picking the root 52 without checking that Q must lie on side AB. (2021 AMC Fall 10B #15)
  • Mixing up which small triangle has which leg ratio (3:4 versus 4:3) when splitting the hypotenuse, or inverting the final ratio to get 35/37. (2017 AMC 10A #21)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.

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