Math Core

Similar triangles are the engine behind most AMC geometry. Whenever you see parallel lines, a right angle with an altitude, or two angles that match, a pair of similar triangles is hiding in the figure, and it hands you a proportion. The skill is spotting the pair fast and matching the vertices correctly.

Spotting similar triangles

Two triangles are similar when their angles match. Two matching angles are enough (AA), because the third angle is then forced. On the AMC, similar pairs almost always come from one of four setups.

A line parallel to a side. If $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ , then $\triangle ADE \sim \triangle ABC$ .
Crossing lines between parallels (an "hourglass"). In trapezoid $ABCD$ with $AB \parallel CD$ , the diagonals meet at $P$ and $\triangle PAB \sim \triangle PCD$ .
The altitude to the hypotenuse. In a right triangle, the altitude to the hypotenuse creates two smaller triangles, and all three triangles are similar.
A shared angle. If $D$ is on side $AC$ of $\triangle ABC$ and $\angle ABD = \angle ACB$ , then $\triangle ABD \sim \triangle ACB$ (they share angle $A$ ).

What similarity gives you

If two triangles are similar with ratio $k$ (every length in one is $k$ times the matching length in the other), then

matching sides, altitudes, medians, inradii and perimeters are all in ratio $k$ ;
areas are in ratio $k^2$ .

Common mistake

Match vertices by angles, not by position on the page. For the shared-angle setup, $\triangle ABD \sim \triangle ACB$ (not $\triangle ABC$ ): $A \leftrightarrow A$ , $B \leftrightarrow C$ , $D \leftrightarrow B$ . Write the correspondence down before writing a proportion.

Parallel lines and trapezoids

A trapezoid with bases 6 (top) and 10 (bottom). The diagonals cross at the dot, which divides each diagonal in the ratio 3 : 5.

Worked example: Trapezoid diagonals

A trapezoid has bases $6$ and $10$ and height $8$ . Its diagonals meet at $P$ . What is the area of the triangle formed by $P$ and the shorter base?

The two triangles formed by $P$ and the bases are similar (an hourglass), with ratio $6 : 10 = 3 : 5$ . Their heights are in the same ratio and add up to $8$ , so they are $3$ and $5$ . The triangle on the short base has area $\tfrac12 \cdot 6 \cdot 3 = 9$ .

A useful by-product: in any trapezoid, the two triangles on the legs (the non-parallel sides) have equal area. If the triangles on the bases have areas $X$ and $Y$ , each leg triangle has area $\sqrt{XY}$ .

Right triangles and the altitude

In right triangle $ABC$ with right angle at $C$ , drop the altitude $CD$ to the hypotenuse. Call $AD = p$ and $DB = q$ . The three similar triangles give

$CD^2 = pq, \qquad AC^2 = p \cdot AB, \qquad BC^2 = q \cdot AB.$

Worked example: The altitude to the hypotenuse

A right triangle has legs $15$ and $20$ . Find the altitude to the hypotenuse and the two pieces it cuts the hypotenuse into.

The hypotenuse is $25$ . The area is $\tfrac12 \cdot 15 \cdot 20 = 150$ , so the altitude is $\dfrac{2 \cdot 150}{25} = 12$ . From $AC^2 = p \cdot AB$ : $p = \dfrac{225}{25} = 9$ , and $q = \dfrac{400}{25} = 16$ . Check: $9 \cdot 16 = 144 = 12^2$ .

Worked example: A square in a triangle

A triangle has base $12$ and height $8$ . A square sits on the base with its top two corners on the other two sides. How long is a side of the square?

Let the side be $s$ . The top of the square cuts off a small triangle at the apex that is similar to the whole triangle. Its base is $s$ and its height is $8 - s$ , so

$\frac{s}{12} = \frac{8 - s}{8} \implies 8s = 96 - 12s \implies s = \frac{24}{5}.$

Worked example: A shared angle

In $\triangle ABC$ , $AB = 6$ . Point $D$ is on $\overline{AC}$ with $AD = 4$ and $\angle ABD = \angle ACB$ . Find $AC$ .

$\triangle ABD \sim \triangle ACB$ with $A \leftrightarrow A$ , $B \leftrightarrow C$ , $D \leftrightarrow B$ . So $\dfrac{AB}{AC} = \dfrac{AD}{AB}$ , which gives $AB^2 = AD \cdot AC$ , so $36 = 4 \cdot AC$ and $AC = 9$ .

Tip

The Angle Bisector Theorem is a close cousin: if the bisector of angle $A$ meets $BC$ at $D$ , then $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ . (Triangles $ABD$ and $ACD$ have the same height from $A$ , and the same height from $D$ to the two sides, so their area ratio is both $\dfrac{BD}{DC}$ and $\dfrac{AB}{AC}$ .)

Practice

Practice 1

In $\triangle ABC$ , a line parallel to $\overline{BC}$ meets $\overline{AB}$ at $D$ and $\overline{AC}$ at $E$ . If $AD = 4$ , $DB = 6$ and $\triangle ADE$ has area $12$ , what is the area of trapezoid $DBCE$ ?

Practice 2

The altitude to the hypotenuse of a right triangle cuts the hypotenuse into segments of lengths $4$ and $9$ . What is the area of the triangle?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A square is drawn inside a right triangle with legs $6$ and $8$ so that one corner of the square is at the right angle and the opposite corner lies on the hypotenuse. What is the side length of the square?

Practice 4

Trapezoid $ABCD$ has bases $AB = 4$ and $CD = 12$ and area $64$ . Its diagonals meet at $P$ . What is the area of $\triangle PBC$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

In $\triangle ABC$ , $AB = 8$ , $BC = 14$ and $CA = 16$ . Point $D$ lies on $\overline{AC}$ so that $\angle ABD = \angle ACB$ . What is $BD + DC$ ?

Practice 6

In $\triangle ABC$ , $AB = 10$ , $AC = 15$ and $BC = 20$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ . What is $BD$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Rectangle $ABCD$ has $AB = 12$ and $BC = 5$ . Point $E$ is the midpoint of $\overline{CD}$ , and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F$ . What is the area of quadrilateral $AFED$ ?

Practice 8

In right triangle $ABC$ with right angle at $C$ , the altitude $\overline{CD}$ is drawn to the hypotenuse. The inradius of $\triangle ACD$ is $3$ and the inradius of $\triangle BCD$ is $4$ . What is the inradius of $\triangle ABC$ ?

Real contest practice

2018 AMC 10A, Problem 9: a figure built from similar isosceles triangles; areas scale by the square of the ratio.
2018 AMC 10A, Problem 24: a midsegment creates similar triangles, combined with the Angle Bisector Theorem.