Similarity

• A ratio compares two quantities by division. The ratio of $6$ to $4$ can be written $6:4$, “$6$ to $4$”, or $\frac{6}{4}=\frac{3}{2}$.
• A proportion is an equation saying two ratios are equal, like $\frac{a}{b}=\frac{c}{d}$.
• To solve a proportion, use cross-multiplication: if $\frac{a}{b}=\frac{c}{d}$, then $a\cdot d = b\cdot c$.
• The scale factor is the number you multiply every length of a small figure by to get the matching length of the larger figure. It is just the ratio of a pair of corresponding sides.

Two polygons are similar ($\sim$) when they have the same shape but not necessarily the same size. This requires both:

• All pairs of corresponding angles are congruent (equal).
• All pairs of corresponding sides are proportional (same scale factor).

A similarity statement lists vertices in matching order. Writing $\triangle ABC \sim \triangle DEF$ tells you $A\!\leftrightarrow\! D$, $B\!\leftrightarrow\! E$, $C\!\leftrightarrow\! F$, so $\angle A\cong\angle D$ and $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$.

You do not need to check everything. For triangles, any one of these guarantees similarity:

• AA (Angle-Angle): two pairs of corresponding angles are congruent. (The third pair is then automatic.)
• SSS$\sim$ (Side-Side-Side): all three pairs of corresponding sides are proportional.
• SAS$\sim$ (Side-Angle-Side): two pairs of corresponding sides are proportional and the included angles (between those sides) are congruent.

When two triangles are similar, corresponding sides are proportional. To find a missing length:

• Write the similarity statement so vertices match.
• Build a proportion using one known pair of sides and the pair that contains the unknown.
• Cross-multiply and solve.

You can find a height you cannot reach by using similar triangles.

• Shadow method: At the same time of day, an object and its shadow form a triangle similar to a nearby object and its shadow (the sun's rays hit at the same angle). So $\dfrac{\text{height}_1}{\text{shadow}_1}=\dfrac{\text{height}_2}{\text{shadow}_2}$.
• Mirror method: Place a mirror on the ground. The angle of sight into the mirror equals the angle back out, creating two similar triangles.

A midsegment of a triangle joins the midpoints of two sides. The Triangle Midsegment Theorem says a midsegment is:

• parallel to the third side, and
• exactly half the length of that third side.

This creates a small triangle similar to the whole triangle with scale factor $\tfrac12$.

If a line is drawn parallel to one side of a triangle and crosses the other two sides, it divides those two sides proportionally.

This happens because the parallel line creates a smaller similar triangle (by AA).

If two similar figures have scale factor $k$ (each length of the big figure is $k$ times the matching length of the small one), then:

• The ratio of perimeters equals the scale factor: $k$.
• The ratio of areas equals the scale factor squared: $k^2$.

Drop the altitude from the right angle of a right triangle to the hypotenuse. It splits the big triangle into two smaller triangles, each similar to the original and to each other (all three share the same angles, by AA). Label the hypotenuse pieces $p$ and $q$, the altitude $h$, and the legs $a$ and $b$.

Matching corresponding sides across the similar triangles gives three “geometric mean” (mean proportional) relations:

In words: the altitude is the geometric mean of the two hypotenuse pieces, and each leg is the geometric mean of its adjacent piece and the whole hypotenuse.

Claim. In $\triangle ABC$, if $\overline{DE}\parallel\overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$, then $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

Proof. Because $\overline{DE}\parallel\overline{BC}$, the parallel lines cut off equal corresponding angles: $\angle ADE\cong\angle ABC$ and $\angle AED\cong\angle ACB$. Triangle $ADE$ and triangle $ABC$ also share $\angle A$. So by AA, $\triangle ADE\sim\triangle ABC$. Similar triangles have proportional corresponding sides, giving

Now flip each ratio to compare whole-to-part, then subtract $1$ from both sides:

Since $AB-AD=DB$ and $AC-AE=EC$, this is exactly $\dfrac{DB}{AD}=\dfrac{EC}{AE}$, and taking reciprocals gives $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. $\blacksquare$

When two figures are similar with scale factor $k$ (every length of the large figure is $k$ times the matching length of the small one), the effect compounds with each dimension:

• Lengths (sides, perimeters, radii, heights) scale by $k^1=k$.
• Areas (surface areas too) scale by $k^2$, because area multiplies two lengths.
• Volumes scale by $k^3$, because volume multiplies three lengths.

Read backwards, this is powerful: if you know the area or volume ratio, take a square or cube root to recover the length scale factor. If $\dfrac{\text{Area}_1}{\text{Area}_2}=\dfrac{9}{25}$, then $k=\sqrt{9/25}=\dfrac{3}{5}$.

Similar triangles often share a vertex or side, so they overlap on the page. Two habits keep them straight:

• Redraw them separately, pulling the small triangle out of the big one and turning it so matching vertices sit in the same positions.
• Use the whole, not the piece. When a small triangle sits inside a big one along a shared side, the big triangle's side is the entire length, not just the leftover part. Mixing a piece with a whole is the most common error.

A shared angle plus a pair of parallel sides (or a pair of equal marked angles) is the usual AA setup that makes the overlap similar.

Claim. If a ray from vertex $A$ bisects $\angle A$ and meets $\overline{BC}$ at $D$, then $D$ divides $\overline{BC}$ in the ratio of the two adjacent sides:

Proof idea. Through $C$ draw a line parallel to the bisector $\overline{AD}$, meeting line $\overline{BA}$ extended at a point $E$. Since $\overline{AD}\parallel\overline{CE}$, the side-splitter theorem in $\triangle BCE$ gives $\dfrac{BD}{DC}=\dfrac{BA}{AE}$. The parallel lines also force $\triangle ACE$ to be isosceles ($\angle AEC$ and $\angle ACE$ equal the two bisected halves of $\angle A$), so $AE=AC$. Substituting yields $\dfrac{BD}{DC}=\dfrac{AB}{AC}$. $\blacksquare$

Sometimes a single similar-triangle setup hides two unknowns — for example an unknown height and an unknown distance. The fix is to write two proportions (or one proportion plus one length equation) so you have as many equations as unknowns, then solve the system.

A classic case: an observer walks toward a tall object, and you know the two viewing distances but neither the object's height nor the observer's starting gap. Set the height as one variable and the hidden distance as another, build a proportion at each observation, and eliminate.