Similar triangles are triangles of the same shape but different size: their angles are respectively equal, and their corresponding sides are proportional. The number showing how many times larger one triangle is than the other is called the similarity ratio \(k\). Three criteria help prove similarity, and the ratios of perimeters and areas follow a simple rule: perimeters relate as \(k\), and areas as \(k^2\). This page covers the definition and notation of similarity with the correct vertex order, all three criteria, similarity of right triangles and the altitude from the right angle, OGE problem walkthroughs on "find the side," and an interactive simulator.
What are similar triangles and the similarity ratio
Two triangles are called similar if two conditions are met simultaneously:
- their angles are respectively equal — each angle of the first triangle corresponds to an equal angle in the second;
- their corresponding sides are proportional — all three ratios yield the same number.
Corresponding sides are sides that lie opposite equal angles. These are the pairs between which the ratio is calculated.
Similarity is denoted by the symbol \(\sim\):
Similarity ratio \(k\) — the identical ratio of corresponding sides:
If \(k > 1\), the second triangle is larger than the first; if \(k < 1\) — it is smaller; at \(k = 1\) the triangles are simply congruent, as congruence is a special case of similarity.
Vertex order in notation is not a formality
The notation \(\triangle ABC \sim \triangle A_1B_1C_1\) is read strictly by position: \(A\) corresponds to \(A_1\), \(B\) to \(B_1\), \(C\) to \(C_1\). This immediately implies which angles are equal (\(\angle A = \angle A_1\)) and which sides are corresponding (\(AB\) and \(A_1B_1\), \(BC\) and \(B_1C_1\), \(AC\) and \(A_1C_1\)).
If you swap letters and write \(\triangle ABC \sim \triangle B_1A_1C_1\), you get a different statement: now \(\angle A = \angle B_1\), and the side corresponding to \(AB\) becomes \(B_1A_1\). You cannot form a proportion based on "similarly positioned" sides — only by positions in the notation. Therefore, first, list the pairs of equal angles and adjust the letter order accordingly.
Three triangle similarity criteria
You don't need to check all six conditions from the definition: similarity is proven by one of the three criteria. Each is a minimal set of data from which the equality of all angles and the proportionality of all sides follow automatically.
How to choose a criterion. Look at what is given in the problem: if it's about angles — the first, if about two sides and the angle between them — the second, if about all three sides — the third.
- First criterion (by two angles) — the workhorse of the school curriculum: almost all OGE problems are solved using it. You don't need to check the third angle; it is equal automatically: the sum of angles in a triangle is always \(180°\). Look for ready-made pairs of equal angles — a common angle of two triangles, vertical angles, alternate interior and corresponding angles with parallel lines, two right angles.
- Second criterion (by two sides and the angle between them) requires the angle to be exactly between these two sides. If the equal angle lies opposite one of them, the criterion does not work, and similarity is not proven.
- Third criterion (by three sides) — when there are no angles in the condition at all, only lengths. List the sides of each triangle in increasing order and divide the smallest by the smallest, the middle by the middle, the largest by the largest: this way, pairs won't get mixed up. Similarity exists only if all three ratios match.
To avoid confusing similarity criteria with congruence criteria, keep the difference in mind: in congruence, sides must coincide, in similarity — be proportional.
→ △ABC ∼ △A1B1C1
Ratio of perimeters and areas of similar triangles
Similarity has two consequences that are asked about most often, and one of them is the main trap of the topic.
Perimeters relate as \(k\). Each side of the second triangle is \(k\) times larger than the corresponding side of the first, so the sum of the sides also grows exactly \(k\) times:
All linear quantities behave the same way — "\(k\) times": altitudes, medians, angle bisectors, radii of inscribed and circumscribed circles, midsegments.
Areas relate as \(k^2\). Area depends on two dimensions at once: both the base and the altitude increase \(k\) times, and the product — \(k \cdot k\) times:
Let \(k = 5\). Then the perimeter is 5 times larger (for example, it was 14 cm — it became 70 cm), and the area — 25 times larger (it was 6 cm² — it became 150 cm²). The difference is huge, and substituting \(k\) instead of \(k^2\) here is the most common mistake on the exam.
Reverse process. If the ratio of areas is given in the condition, the similarity ratio is extracted by the root: \(k = \sqrt{S_2 : S_1}\). For example, areas relate as \(16 : 1\), which means sides relate as \(4 : 1\).
| k | Perimeters P₂ : P₁ | Areas S₂ : S₁ |
|---|---|---|
| 2 | 2 : 1 | 4 : 1 |
| 3 | 3 : 1 | 9 : 1 |
| 1.5 | 1.5 : 1 | 2.25 : 1 |
| any | k : 1 | k² : 1 |
Similarity of right triangles and the altitude from the right angle
For right triangles, the first criterion simplifies: their right angles are equal in advance, so equality of one acute angle is sufficient for the triangles to be similar. The criterion by two legs follows from this: if the legs of one are proportional to the legs of the other, the triangles are similar (this is the second criterion, since the angle between the legs is a right angle).
The most useful configuration of the entire topic is the altitude drawn from the vertex of the right angle. It divides the triangle into two, and all three resulting triangles are similar to each other.
Why so: triangle \(AHC\) has a right angle at \(H\) and a common angle \(A\) with the large triangle, so \(\triangle AHC \sim \triangle ACB\). Triangle \(CHB\) has a right angle at \(H\) and a common angle \(B\), so \(\triangle CHB \sim \triangle ACB\). Hence, the two small triangles are also similar to each other.
From these similarities, formulas for the geometric mean are derived (they are called proportional segments in a right triangle):
- \(CH^2 = AH \cdot HB\) — the altitude is the geometric mean of the segments into which it divides the hypotenuse;
- \(AC^2 = AH \cdot AB\) and \(BC^2 = HB \cdot AB\) — a leg is the geometric mean of the entire hypotenuse and its projection onto it.
The main thing is not to confuse which segment is "its own": for leg \(AC\), the projection is \(AH\), for leg \(BC\), the projection is \(HB\). The projection always adjoins the same end of the hypotenuse as the leg.
OGE problems on similarity: how to find the side
Almost all similarity problems are solved using one algorithm:
- Find a pair of equal angles — a common angle, vertical, alternate interior with parallel lines, or two right angles.
- Write down similarity in the correct vertex order — so that equal angles are in the same positions.
- Form a proportion from corresponding sides (take sides that are in the same places in the notation).
- Express the unknown and calculate; at the end, check the answer for sense — the side of the smaller triangle must be smaller.
There are few typical configurations in the OGE, and they are worth recognizing at first sight:
- A line parallel to a side. The line cuts off a small triangle from the triangle, similar to the original one. Key subtlety: the ratio of whole sides from the common vertex (\(BM : BA\)) goes into the proportion, not the segments (\(BM : MA\)).
- "Butterfly." Two segments intersect, and their ends are connected by parallel lines — two similar triangles are formed, rotated relative to each other.
- Altitude from the right angle — the configuration from the previous section; in the OGE, it is used to find an altitude, a leg, or a segment of the hypotenuse.
- Shadow and height of an object. Sun rays are parallel, so an object with its shadow and a tree with its shadow form similar right triangles.
- Midsegment — a special case with \(k = 2\): it is parallel to the side and half its length.
angle B is common, angles at M and A are corresponding
angles at O are vertical, angles A and C are alternate interior
Examples of solving triangle similarity problems
Example 1. Find sides by similarity ratio
Problem. \(\triangle MNK \sim \triangle PQR\). It is known that \(MN = 7\), \(NK = 11\), \(MK = 9\) and \(PQ = 21\). Find \(QR\) and \(PR\).
Step 1. Analyze the notation by position: \(M \leftrightarrow P\), \(N \leftrightarrow Q\), \(K \leftrightarrow R\). Thus, the corresponding pairs are: \(MN\) and \(PQ\), \(NK\) and \(QR\), \(MK\) and \(PR\).
Step 2. Calculate the ratio using the only pair where both sides are known:
Step 3. The other sides of the second triangle are 3 times larger than the corresponding ones:
Answer: \(QR = 33\), \(PR = 27\). Verification: \(33 : 11 = 27 : 9 = 3\) — all ratios matched.
Example 2. Prove similarity by the second criterion
Problem. In triangles \(ABC\) and \(DEF\), it is known: \(AB = 5\), \(AC = 9\), \(DE = 15\), \(DF = 27\), and angles \(A\) and \(D\) are equal. Prove that the triangles are similar.
Step 1. Check the main condition of the second criterion: angle \(A\) is enclosed between sides \(AB\) and \(AC\), angle \(D\) — between \(DE\) and \(DF\). Angles between the given sides — the criterion is applicable.
Step 2. Calculate the ratios of corresponding sides:
Step 3. Two pairs of sides are proportional with the same ratio \(k = 3\), the angles between them are equal — by the second criterion \(\triangle ABC \sim \triangle DEF\).
Answer: triangles are similar, \(k = 3\). Note: if angles \(B\) and \(D\) were equal, the criterion would not work — angle \(B\) does not lie between \(AB\) and \(AC\).
Example 3. A line parallel to a side (OGE problem)
Problem. In triangle \(ABC\), point \(M\) lies on side \(AB\), point \(N\) — on side \(BC\), and \(MN \parallel AC\). It is known that \(BM = 4\), \(MA = 12\), \(MN = 5\). Find \(AC\).
Step 1. Since \(MN \parallel AC\), angles \(BMN\) and \(BAC\) are equal as corresponding, and angle \(B\) is common. By the first criterion \(\triangle BMN \sim \triangle BAC\) (vertex order: \(M \leftrightarrow A\), \(N \leftrightarrow C\)).
Step 2. Find the full side from the common vertex: \(BA = BM + MA = 4 + 12 = 16\). This is the main subtlety — \(BA\) goes into the proportion, not \(MA\).
Step 3. Form a proportion from corresponding sides:
Step 4. Hence \(AC = 5 \cdot 4 = 20\).
Answer: \(AC = 20\). The similarity ratio here is 4: the large triangle is four times larger than the cut-off one.
Example 4. Perimeters and areas
Problem. The perimeters of similar triangles are 18 cm and 45 cm. The area of the smaller triangle is 16 cm². Find the area of the larger one.
Step 1. Perimeters relate as the similarity ratio:
Step 2. Areas relate as the square of the ratio: \(k^2 = 2.5^2 = 6.25\).
Step 3. Multiply the area of the smaller triangle by \(k^2\):
Answer: 100 cm². If we had multiplied by \(k = 2.5\), we would have obtained 40 cm² — a typical mistake that causes the loss of the entire point.
Example 5. Altitude from the vertex of the right angle
Problem. In a right triangle \(ABC\) with a right angle at vertex \(C\), an altitude \(CH\) is drawn to the hypotenuse. It is known that \(AH = 4\) cm, \(HB = 9\) cm. Find the altitude \(CH\) and the area of the triangle.
Step 1. The altitude from the right angle is the geometric mean of the hypotenuse segments, because \(\triangle AHC \sim \triangle CHB\):
Step 2. Extract the root: \(CH = 6\) cm.
Step 3. Hypotenuse \(AB = AH + HB = 4 + 9 = 13\) cm. Calculate the area through the hypotenuse and the altitude drawn to it:
Answer: \(CH = 6\) cm, \(S = 39\) cm².
Example 6. Tree height by shadow
Problem. A vertical pole 2 m high casts a shadow 1.5 m long. At the same moment, the tree's shadow is 9 m. What is the tree's height?
Step 1. Sun rays are parallel, so the angle of incidence for the pole and the tree is the same. The pole with its shadow and the tree with its shadow form right triangles with an equal acute angle — meaning they are similar.
Step 2. Form a proportion "height to its shadow":
Step 3. Express the unknown height:
Answer: 12 m. Verification by sense: the tree's shadow is 6 times longer than the pole's shadow, so the tree is 6 times higher: \(2 \cdot 6 = 12\) m.
Common mistakes in triangle similarity problems
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Assume that the areas of similar triangles relate as $k$: at $k = 3$, they increase the area threefold.
Areas relate as the square of the ratio: at \(k = 3\), the area is \(3^2 = 9\) times larger. Only linear quantities relate as \(k\) — sides, perimeter, altitudes, medians. Hint rule: area is measured in square units, so the ratio enters it squared.
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Ignore vertex order: from the notation $\triangle ABC \sim \triangle KLM$, they form the proportion $AB : LM$.
Vertex correspondence is defined by positions in the notation: \(A \leftrightarrow K\), \(B \leftrightarrow L\), \(C \leftrightarrow M\). The side corresponding to \(AB\) will be \(KL\), and to \(BC\) — \(LM\). Before writing the proportion, list the vertex pairs in a column — this takes five seconds and eliminates most mistakes.
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Prove similarity by one equal angle or by two proportional sides without an angle.
One equal angle is not enough: two triangles can have a matching angle while having completely different shapes. A full criterion is needed — two angles, or two sides and the angle between them, or all three sides.
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In the second criterion, they take an angle that does not lie between the given sides.
The second criterion works only with the angle between the two given sides. If \(AB\), \(AC\) are known and angle \(B\) is equal, the criterion is inapplicable — look for another path: a pair of equal angles or a third side.
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Reverse the ratio: they find $k$ as "smaller side to larger," and then multiply by it to get the side of the larger triangle.
Agree with yourself on the direction: if \(k\) is the ratio of the second triangle's sides to the first, then the first triangle's sides are multiplied by \(k\), and the second triangle's sides are divided by \(k\). Final verification by sense: the side of the smaller triangle must be smaller.
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In a problem with a line parallel to a side, they form a proportion from segments: $BM : MA = MN : AC$.
Triangles \(BMN\) and \(BAC\) are similar, and their corresponding sides are \(BM\) and \(BA\) (the whole side from the common vertex), not the segment \(MA\). First find \(BA = BM + MA\), and only then form the proportion.
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Confuse similarity with congruence and write in the answer that the sides are simply equal.
Congruence is a special case of similarity at \(k = 1\). If \(k \ne 1\), only the angles of the triangles match, and the sides differ by a factor of \(k\). The criteria are also different: for congruence, it's SAS, ASA, SSS with coinciding sides, for similarity — proportional sides.
Questions and answers
Which triangles are called similar?
Triangles are called similar if their angles are respectively equal and their corresponding sides are proportional. Simply put, these are triangles of the same shape but different size. Similarity is denoted by the symbol ∼: △ABC ∼ △A₁B₁C₁.
What is the triangle similarity ratio?
The similarity ratio k is a number equal to the ratio of corresponding sides of similar triangles: k = A₁B₁ : AB = B₁C₁ : BC = A₁C₁ : AC. It shows how many times larger one triangle is than the other. At k = 1, the triangles are congruent.
How many triangle similarity criteria are there and what are they?
There are three criteria. The first — by two angles: two angles of one triangle are equal to two angles of the other. The second — by two sides and the angle between them: two sides are proportional and the angles between them are equal. The third — by three sides: all three pairs of corresponding sides are proportional.
How do the areas of similar triangles relate?
The areas of similar triangles relate as the square of the similarity ratio: S₂ : S₁ = k². For example, at k = 4, the area is 16 times larger. Conversely: if areas relate as 25 : 1, then the sides relate as 5 : 1.
How do the perimeters of similar triangles relate?
Perimeters relate as the similarity ratio: P₂ : P₁ = k. All linear elements behave the same way — altitudes, medians, angle bisectors, midsegments, radii of inscribed and circumscribed circles.
How does triangle similarity differ from congruence?
In congruent triangles, both angles and sides match, so the figures can be superimposed on each other. In similar ones, only the angles match, and the sides are proportional with ratio k. Congruence is a special case of similarity at k = 1.
How to prove the similarity of right triangles?
Equality of one acute angle is sufficient: their right angles are already equal, so the first criterion works. The second method is the proportionality of legs: the angle between the legs is a right angle, so the second similarity criterion is applicable.
Why does the altitude from the right angle give three similar triangles?
The altitude CH from the vertex of the right angle C divides triangle ABC into triangles AHC and CHB. AHC has a right angle at H and a common angle A with ABC, CHB has a right angle at H and a common angle B, so both are similar to the original, and thus to each other. From this, the formulas CH² = AH · HB, AC² = AH · AB, and BC² = HB · AB are derived.
In which grade is triangle similarity taught?
In grade 8 in geometry lessons — right after the topic on proportional segments. There, three similarity criteria, the ratio of areas, and the application of similarity to a right triangle are studied. Later, the topic is used in grade 9 and in OGE problems.
