Similar triangles: the ratio tricks that save time

Similar-triangle questions are rarely slow because the similarity is hard to see. They are slow because of what people do once they have seen it: write a proportion with four lengths in it, cross-multiply, match the wrong sides, and redo the line. On a timed paper that costs two or three minutes a problem, and the problem was worth thirty seconds.

These are the habits that remove the bookkeeping. None of them is new mathematics. They are just the ways strong students write the same fact down faster.

Trick 1: write the correspondence in order, once

Before any lengths, write the similarity statement with the vertices matched by angle. If $\angle A = \angle D$ and $\angle B = \angle E$, write $\triangle ABC \sim \triangle DEF$, in that order, and from then on never think about which side matches which. Position does it for you: the first two letters give $AB \leftrightarrow DE$, the last two $BC \leftrightarrow EF$, first and last $AC \leftrightarrow DF$.

Most wrong answers in this topic are not wrong similarity. They are a correct similarity read with the sides crossed. Writing the vertices in angle order is the whole cure, and it takes five seconds.

Trick 2: one number, not a proportion

A proportion like $\frac{x}{12} = \frac{8}{15}$ has four slots and invites you to put a length in the wrong one. Replace it with a single scale factor $k$: every length in the small triangle is $k$ times its partner in the big one. Find $k$ from the one pair of sides you know, then multiply. There is nothing to cross.

The slip that $k$ exposes is the commonest one in the topic. If a point $D$ on $AB$ splits it as $AD : DB = 2 : 3$, the scale factor from $\triangle ABC$ down to the small triangle at $A$ is

$$k = \frac{AD}{AB} = \frac{2}{2+3} = \frac25,$$

not $\frac23$. Ratios along a side are part-to-part; scale factors are part-to-whole. Convert before you use it.

Trick 3: areas scale by k², and the leftover is 1 − k²

Lengths scale by $k$, areas by $k^2$, volumes by $k^3$. Everybody knows it, and it is still the single most expensive mistake here, because the wrong answer ($k$ instead of $k^2$) is always one of the options.

The time saver is the complement. A line parallel to $BC$ that cuts off the triangle $ADE$ with $k = \frac25$ leaves that triangle with $\frac{4}{25}$ of the area, so the trapezoid $DBCE$ has $1 - \frac{4}{25} = \frac{21}{25}$ of it. You never compute a height. The same move says the midsegment triangle has a quarter of the area, since $k = \frac12$, and the strip below it the other three quarters.

Trick 4: memorise the three shapes, not the theorem

Similarity on contest papers comes in three shapes far more often than any other. Learn what each one outputs and you skip the derivation every time.

The nested triangle. A line parallel to one side cuts off a smaller copy at the opposite vertex. Output: scale factor $k = \frac{AD}{AB}$, the parallel side is $k$ times the base, area $k^2$. This is the shape behind Trick 3.

The bowtie. Two parallel segments joined by two crossing lines, most often a trapezoid with its diagonals. The two triangles touching the parallel sides are similar, with $k$ equal to the ratio of the parallel sides, and the crossing point cuts each diagonal in that ratio. In a trapezoid with parallel sides $6$ and $10$, the diagonals cut each other $3 : 5$. The four triangles then have areas in the ratio

$$9 : 15 : 25 : 15,$$

because the two side triangles are always equal, and each is the geometric mean of the other two. That makes the total $(3+5)^2 = 64$ parts, and gives a formula worth knowing: the trapezoid's area is $\left(\sqrt{T_1} + \sqrt{T_2}\right)^2$, where $T_1$ and $T_2$ are the areas of the two triangles on the parallel sides.

The right-triangle altitude. Drop the altitude from the right angle $A$ to the hypotenuse, meeting it at $H$. All three triangles in the picture are similar, and the outputs are

• $AH^2 = BH \cdot HC$
• $AB^2 = BH \cdot BC$
• $AC^2 = HC \cdot BC$

With $BH = 4$ and $HC = 9$: $AH = 6$, $AB = \sqrt{4 \cdot 13} = 2\sqrt{13}$ and $AC = \sqrt{9 \cdot 13} = 3\sqrt{13}$. Three lengths, no Pythagoras, and Pythagoras confirms them afterwards: $52 + 117 = 169 = 13^2$.

Where the shapes hide

The reason these three are worth memorising is that they turn up inside problems that do not mention similarity at all.

The medians cut each other 2 : 1 is a bowtie. Join the midpoints of $AB$ and $AC$: that midsegment is parallel to $BC$ and half its length, so the two medians from $B$ and $C$ cross between two parallel segments with ratio $1 : 2$. The crossing point divides each of them $2 : 1$, and the centroid fact is that bowtie with $k = \frac12$.

The angle bisector theorem is a nested triangle in disguise. Through $C$, draw the line parallel to the bisector $AD$, and extend $BA$ to meet it at $E$. The parallel gives two equal angles at $E$ and $C$, so triangle $ACE$ is isosceles with $AE = AC$. Now $AD$ is a parallel cut in triangle $BEC$, so $\frac{BD}{DC} = \frac{BA}{AE} = \frac{AB}{AC}$. A theorem that looks like a fact about angles is one parallel line away from a scale factor.

That is the general lesson. When a problem hands you a ratio along a side and nothing seems to produce it, draw the parallel line that would make a nested triangle or a bowtie. The ratio usually appears in the first line after it.

The thirty-second routine

When you recognise a similarity, do these in order and write nothing else until they are done:

1. Write the similarity with the vertices in angle order.
2. Turn the one known pair of sides into a single $k$, converting part-to-part ratios to part-to-whole.
3. If the question asks about an area, square $k$ before you do anything else. If it asks about a leftover region, take $1 - k^2$.

To build the habit, take a page of length-chasing problems and for each one do only these three steps: name the similar pair, write $k$, and say whether the answer needs $k$, $k^2$ or $1 - k^2$. You will get through a page in the time a single full solution takes, and the speed you are short of lives in exactly those three lines.

The similarity and ratios page has the companions in their shortest form, including where the topic sits in the ladder, and the problem archive has the two proofs above in full. For a cold start each day, the daily problem posts a new one, no account needed.