Similarity and Theorems on Similar Triangles (Similarity Theorems)
We often see figures that have the same shape but different sizes. For example, the same photograph may appear in stamp, passport and postcard sizes. Their size changes, but their shape remains the same. Such figures are called similar figures.
The main aim of this topic is to understand when two figures, especially two triangles, are similar and what proportional or geometric relations follow from similarity.
📚 Basic idea of similar figures
👉 Two congruent figures are always similar.
👉 Two similar figures need not be congruent.
👉 All squares are similar to one another.
👉 All circles are similar to one another.
👉 All equilateral triangles are similar to one another.
👉 A square and a rhombus are not always similar because their corresponding angles need not be equal.
📚 Conditions for similar polygons
Two polygons having the same number of sides are similar if:
- their corresponding angles are equal, and
- their corresponding sides are proportional.
Suppose ABCD and A'B'C'D' are similar quadrilaterals. Then,
∠A = ∠A', ∠B = ∠B', ∠C = ∠C', ∠D = ∠D'and
AB/A'B' = BC/B'C' = CD/C'D' = DA/D'A'.
📚 Similar triangles
If two triangles are similar,
👉 their corresponding angles are equal, and
👉 their corresponding sides are proportional.
Suppose △ABC ~ △DEF. Then,
∠A = ∠D, ∠B = ∠E, ∠C = ∠Fand
AB/DE = BC/EF = AC/DF.
📚 Symbol of similarity
The symbol ~ is used for similarity. Thus,
△ABC ~ △DEF.
The order of letters is important. It shows that A ↔ D, B ↔ E and C ↔ F are corresponding vertices.
📚 Dividing a line segment in a given ratio
Let P lie on the segment AB. Then P divides AB internally in the ratio AP : PB.
👉 If P is the midpoint of AB, then AP/PB = 1.
👉 If P lies nearer to A, usually AP/PB < 1.
👉 If P lies nearer to B, usually AP/PB > 1.
If the point lies on the extension of the line segment, the ratio represents external division.
⭐ Theorem 43: Basic Proportionality Theorem (BPT) or Thales' Theorem
A line drawn parallel to one side of a triangle divides the other two sides, or their extensions, in the same ratio.
| Given: | In △ABC, DE ∥ BC. The line DE meets AB and AC at D and E respectively. |
| To prove: | AD/DB = AE/EC. |
| Construction: | Join B,E and C,D. Draw DM ⟂ AC and EN ⟂ AB. |
| Proof: | Taking AD and DB as bases, triangles ADE and BDE have the same altitude EN. |
| ∴ | area(△ADE) = 1/2 × AD × EN and area(△BDE) = 1/2 × DB × EN. |
| ∴ | area(△ADE)/area(△BDE) = AD/DB. ...(i) |
| Similarly, | Taking AE and EC as bases, triangles ADE and DEC have the same altitude DM. |
| ∴ | area(△ADE)/area(△DEC) = AE/EC. ...(ii) |
| But, | △BDE and △DEC are on the same base DE and between the same parallels DE and BC. Hence their areas are equal. |
| ∴ | From (i) and (ii), AD/DB = AE/EC. [Proved] |
📚 Two useful forms of BPT
If DE ∥ BC, then from AD/DB = AE/EC, we also get
AB/DB = AC/ECand
AD/AB = AE/AC.
Choose the form that best matches the information given in a problem.
⭐ Theorem 44: Converse of the Basic Proportionality Theorem
If a line divides two sides of a triangle, or their extensions, in the same ratio, then the line is parallel to the third side.
| Given: | A line meets AB and AC of △ABC at D and E such that AD/DB = AE/EC. |
| To prove: | DE ∥ BC. |
| Construction: | Join B,E and C,D. Draw DM ⟂ AC and EN ⟂ AB. |
| Proof: | As in the previous theorem, area(△ADE)/area(△BDE) = AD/DB and area(△ADE)/area(△DEC) = AE/EC. |
| Since, | AD/DB = AE/EC. [Given] |
| ∴ | area(△BDE) = area(△DEC). |
| Therefore, | Triangles BDE and DEC are on the same base DE and have equal areas, so their third vertices B and C lie on a line parallel to DE. |
| ∴ | DE ∥ BC. [Proved] |
📚 Easy way to remember BPT and its converse
👉 Parallel lines given → find ratios.
DE ∥ BC ⇒ AD/DB = AE/EC👉 Equal ratios given → prove parallel lines.
AD/DB = AE/EC ⇒ DE ∥ BC⭐ Theorem 45: Equiangular triangles have proportional corresponding sides
| Given: | △ABC and △DEF are equiangular: ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. |
| To prove: | AB/DE = BC/EF = AC/DF. |
| Construction: | On DE, take DP = AB, and on DF, take DQ = AC. Join PQ. |
| Proof: | In △ABC and △DPQ, AB = DP, AC = DQ and ∠A = ∠D. |
| ∴ | △ABC ≅ △DPQ. [SAS congruence] |
| Hence, | ∠B = ∠P. But ∠B = ∠E, so ∠P = ∠E and therefore PQ ∥ EF. |
| Using BPT, | DP/DE = DQ/DF. Since DP = AB and DQ = AC, AB/DE = AC/DF. ...(i) |
| Similarly, | AB/DE = BC/EF. ...(ii) |
| ∴ | AB/DE = BC/EF = AC/DF. [Proved] |
📚 Result: AA or AAA similarity criterion
If the corresponding angles of two triangles are equal, the triangles are similar. In practice, equality of two corresponding angles is enough because the third angles must also be equal.
∠A = ∠D, ∠B = ∠E ⇒ △ABC ~ △DEF.
⭐ Theorem 46: If corresponding sides are proportional, corresponding angles are equal
| Given: | AB/DE = BC/EF = AC/DF for △ABC and △DEF. |
| To prove: | ∠A = ∠D, ∠B = ∠E, ∠C = ∠F; hence △ABC ~ △DEF. |
| Construction: | On DE, take DP = AB, and on DF, take DQ = AC. Join PQ. |
| Proof: | From AB/DE = AC/DF, with AB = DP and AC = DQ, we get DP/DE = DQ/DF. |
| ∴ | By converse BPT, PQ ∥ EF. Hence ∠P = ∠E and ∠Q = ∠F. |
| Again, | From PQ ∥ EF, DP/DE = PQ/EF. As DP = AB, AB/DE = PQ/EF. ...(i) |
| But, | AB/DE = BC/EF. [Given] ...(ii) |
| ∴ | From (i) and (ii), PQ = BC. |
| Now, | In △ABC and △DPQ, AB = DP, BC = PQ and AC = DQ. |
| ∴ | △ABC ≅ △DPQ. [SSS congruence] |
| Therefore, | ∠A = ∠D, ∠B = ∠P = ∠E and ∠C = ∠Q = ∠F. |
| Hence, | △ABC ~ △DEF. [Proved] |
📚 Result: SSS similarity criterion
If the three pairs of corresponding sides of two triangles are proportional, the triangles are similar.
AB/DE = BC/EF = AC/DF ⇒ △ABC ~ △DEF.
⭐ Theorem 47: SAS similarity criterion
If one pair of corresponding angles is equal and the two pairs of sides including those angles are proportional, the triangles are similar.
| Given: | In △ABC and △DEF, ∠A = ∠D and AB/DE = AC/DF. |
| To prove: | △ABC ~ △DEF. |
| Proof: | Compare the triangles so that the equal angles ∠A and ∠D coincide and the sides AB, AC lie along DE, DF respectively. Because the including sides are proportional, the corresponding division points are in the same ratio. |
| ∴ | By the converse of BPT, the joining line through those points is parallel to the third side. |
| Hence, | ∠B = ∠E and ∠C = ∠F. |
| Therefore, | △ABC ~ △DEF. [Proved] |
📚 Result: SAS similarity criterion
∠A = ∠D and AB/DE = AC/DF ⇒ △ABC ~ △DEF.
📚 When are two triangles similar?
The three main similarity criteria are:
👉 AA/AAA: corresponding angles are equal.
👉 SSS: the three pairs of corresponding sides are proportional.
👉 SAS: one pair of corresponding angles is equal and the sides including those angles are proportional.
If any one of these criteria is correctly satisfied, the two triangles are similar.
⭐ Theorem 48: Similarity in a right triangle when an altitude is drawn to the hypotenuse
If a perpendicular is drawn from the right-angle vertex of a right triangle to its hypotenuse, the two smaller triangles formed are similar to the original triangle and also to each other.
| Given: | △ABC is right-angled at A, and AD ⟂ BC. |
| To prove: | (i) △DBA ~ △ABC, (ii) △DAC ~ △ABC, and (iii) △DBA ~ △DAC. |
| Proof: | In △DBA and △ABC, ∠BDA = ∠BAC = 90° and ∠ABD = ∠CBA. |
| ∴ | △DBA ~ △ABC by AA similarity. [(i) Proved] |
| Again, | In △DAC and △ABC, ∠ADC = ∠BAC = 90° and ∠ACD = ∠BCA. |
| ∴ | △DAC ~ △ABC by AA similarity. [(ii) Proved] |
| Therefore, | Since both smaller triangles are similar to △ABC, △DBA ~ △DAC. [(iii) Proved] |
📚 Three important relations from Theorem 48
For the same figure,
AB² = BC × BD AD² = BD × CDAC² = BC × CD.
The second relation shows that the altitude AD is the mean proportional between the two parts BD and CD of the hypotenuse.
📚 Solved examples
Example 1: Using BPT
In △ABC, DE ∥ BC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, find EC.
By BPT,
AD/DB = AE/EC 4/6 = 5/EC 4EC = 30Therefore,
EC = 7.5 cm.
Example 2: Identifying similar triangles by SSS
The sides of two triangles are 3 cm, 4 cm, 5 cm and 6 cm, 8 cm, 10 cm.
3/6 = 4/8 = 5/10 = 1/2.
So the three pairs of corresponding sides are proportional. Therefore, the triangles are similar by the SSS similarity criterion.
Example 3: Using similarity to find a side
Suppose △ABC ~ △DEF, where AB = 6 cm, DE = 9 cm and BC = 8 cm. Find EF.
Since corresponding sides are proportional,
AB/DE = BC/EF 6/9 = 8/EF 6EF = 72Therefore,
EF = 12 cm.
✍️ Practice problems
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In △ABC, DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, find EC.
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A line cuts AB and AC of △ABC at D and E. If AD = 4 cm, DB = 6 cm, AE = 6 cm and EC = 9 cm, show that DE ∥ BC.
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Two triangles have angles 45°, 65°, 70° and 45°, 65°, 70°. Which similarity criterion proves them similar?
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The sides of two triangles are 5, 7, 8 and 10, 14, 16. Are the triangles similar?
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In two triangles, one included angle is 60°. The sides around that angle are 4 cm, 6 cm in the first triangle and 10 cm, 15 cm in the second. Are the triangles similar?
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In a right triangle, an altitude to the hypotenuse divides it into segments 9 cm and 16 cm. Find the altitude.
Answers
- EC = 7.5 cm
- AD/DB = 4/6 = 2/3 and AE/EC = 6/9 = 2/3. Hence DE ∥ BC by converse BPT.
- AA/AAA similarity.
- Yes, by SSS because 5/10 = 7/14 = 8/16 = 1/2.
- Yes, by SAS because 4/10 = 6/15 = 2/5 and the included angles are equal.
- AD² = 9 × 16 = 144, so AD = 12 cm.
⚠️ Common mistakes
👉 Mistake: Assuming that similar figures must be congruent.
Correct: Congruent figures are always similar, but similar figures can have different sizes.
👉 Mistake: Thinking equal corresponding angles alone are enough for any two polygons to be similar.
Correct: For polygons in general, corresponding angles must be equal and corresponding sides proportional.
👉 Mistake: Writing AD/AE = DB/EC directly from DE ∥ BC.
Correct: Match corresponding segments correctly: AD/DB = AE/EC.
👉 Mistake: Ignoring the order of vertices while writing similar triangles.
Correct: In △ABC ~ △DEF, the correspondence is A ↔ D, B ↔ E, C ↔ F.
👉 Mistake: Using any two side ratios in SAS similarity.
Correct: The proportional sides must be the sides including the equal angle.
👉 Mistake: Using BPT and converse BPT in the same way.
Correct: Parallel lines given → use BPT for ratios. Equal ratios given → use converse BPT to prove parallel lines.
📚 Important points to remember
👉 Similar figures have the same shape, but their sizes may be equal or different.
👉 Congruent figures are always similar, but similar figures are not always congruent.
👉 Similar polygons have equal corresponding angles and proportional corresponding sides.
👉 DE ∥ BC ⇒ AD/DB = AE/EC. This is the Basic Proportionality Theorem.
👉 AD/DB = AE/EC ⇒ DE ∥ BC. This is the converse of BPT.
👉 In AA/AAA similarity, corresponding angles are equal.
👉 In SSS similarity, all three pairs of corresponding sides are proportional.
👉 In SAS similarity, one pair of corresponding angles is equal and the including sides are proportional.
👉 Drawing an altitude from the right-angle vertex to the hypotenuse creates three mutually similar triangles.
👉 In that figure, AB² = BC × BD, AD² = BD × CD and AC² = BC × CD.