Similar & Congruent Triangles

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 $k$, and the areas are in ratio $k^2$. Example: if the sides are $3$ times as long, the area is $9$ times as big.
• A line parallel to one side. It cuts the other two sides in the same proportion. Example: in triangle $ABC$ with $D$ on $AB$, $E$ on $AC$, and $DE\parallel BC$: if $AD = 2$, $DB = 3$, and $AE = 4$, then $\frac{AD}{DB} = \frac{AE}{EC}$ gives $EC = 6$. 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 $30$-$60$-$90$ triangle with short leg $5$ is not congruent to one with long leg $5$), and HL (hypotenuse and a leg, for right triangles).
• The altitude to the hypotenuse. In a right triangle with legs $a$, $b$ and hypotenuse $c$, the altitude $h$ to the hypotenuse makes three similar right triangles. If it cuts the hypotenuse into $p$ (next to leg $a$) and $q$ (next to leg $b$), then $h^2 = pq$, $a^2 = pc$, and $b^2 = qc$. Example: in the $3$-$4$-$5$ triangle, $p = \frac95$ and $q = \frac{16}5$, so $h^2 = \frac{144}{25}$ and $h = \frac{12}{5}$; also $3^2 = \frac95\cdot5$ and $4^2 = \frac{16}{5}\cdot5$.

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.
• "$DE\parallel BC$ splits triangle $ABC$ into two equal areas; find $\frac{AD}{AB}$." Triangle $ADE$ has half the area, so $k^2 = \frac12$ and $k = \frac{\sqrt2}{2}$: the $\sqrt2$ comes from area scaling by $k^2$.

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: $\triangle ABE \sim \triangle FCE$ gives $\frac{AB}{FC} = \frac{BE}{CE}$.
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 $ABCD$ has side length $6$. Point $E$ lies on side $BC$ with $BE = 2$. Line $AE$ meets line $DC$, extended past $C$, at $F$. What is the area of triangle $ADF$?

(A) $36$ (B) $42$ (C) $48$ (D) $54$ (E) $60$

Solution. Step 1. Since $AB\parallel DF$, the alternate angles $\angle BAE$ and $\angle CFE$ are equal. The vertical angles at $E$ are equal too. So $\triangle ABE \sim \triangle FCE$, with $A$, $B$, $E$ matching $F$, $C$, $E$.

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

Step 3. Triangle $ADF$ has a right angle at $D$, base $DF = 18$, and height $AD = 6$. Its area is $\frac12\cdot 18\cdot 6 = 54$.

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

Pitfalls

• Pairing sides that don't match, like $\frac{AB}{CE}$ instead of $\frac{AB}{FC}$. Match sides by the angles across from them: $AB$ and $FC$ are both across from the angle at $E$.
• Forgetting to square the ratio for areas, or to take the square root going from areas back to lengths. Example: sides in ratio $2:3$ mean areas in ratio $4:9$.
• 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)