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Two polygons with the same number of sides are similar if
Both conditions are needed. A square and a rhombus of equal sides have proportional sides but unequal angles; a square and a rectangle have equal angles but sides not in one ratio. Neither pair is similar.
ΔABC ~ ΔDEF (read 'is similar to') means A ↔ D, B ↔ E, C ↔ F. The order matters: it tells you which angles are equal and which sides are in proportion.
Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. Triangles with equal corresponding angles are called equiangular; for these, the sides are automatically in proportion.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Join BE and CD, and draw EM ⊥ AB, DN ⊥ AC.
Triangles BDE and DEC have the same base DE and lie between the parallels DE and BC, so their areas are equal. Hence AD/DB = AE/EC.
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
If DE ∥ BC in ΔABC, each of these holds:
| Criterion | What you need to show | Theorem |
|---|---|---|
| AAA | all three pairs of corresponding angles equal | 6.3 |
| AA | two pairs of corresponding angles equal (the third then follows) | corollary of 6.3 |
| SSS | all three pairs of corresponding sides in the same ratio | 6.4 |
| SAS | one angle equal and the sides including it in the same ratio | 6.5 |
Two sides proportional with a non-included angle equal (SSA) does not guarantee similarity.
| Result | Statement |
|---|---|
| Similar polygons | corresponding angles equal and sides proportional |
| BPT (Thales) | DE ∥ BC ⇒ AD/DB = AE/EC |
| Converse of BPT | AD/DB = AE/EC ⇒ DE ∥ BC |
| AAA / AA | equal angles ⇒ similar |
| SSS | all sides proportional ⇒ similar |
| SAS | one angle equal, including sides proportional ⇒ similar |
Try each problem first, then open the solution to check your working.
In ΔABC, DE ∥ BC with D on AB and E on AC. AD = 4 cm, DB = 6 cm and AE = 5 cm. Find EC.
Given: AD = 4, DB = 6, AE = 5
Use: AD/DB = AE/EC
4/6 = 5/EC ⇒ EC = 30/4 = 7.5
Answer: EC = 7.5 cm
In ΔABC, DE ∥ BC. If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, find x.
Use: AD/DB = AE/EC
x/(x − 2) = (x + 2)/(x − 1)
x² − x = x² − 4 ⇒ x = 4
Answer: x = 4
In ΔABC, DE ∥ BC with AD/DB = 3/5. If AC = 16 cm, find AE.
Given: AD : DB = 3 : 5
Use: AE/AC = AD/AB
AD/AB = 3/(3 + 5) = 3/8
AE = 3/8 × 16 = 6
Answer: AE = 6 cm
D and E are points on AB and AC of ΔABC with AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm. Is DE ∥ BC?
Use: Converse of BPT
AD/DB = 6/9 = 2/3; AE/EC = 8/12 = 2/3
The ratios are equal.
Answer: Yes, DE ∥ BC
In ΔPQR, ST ∥ QR with S on PQ and T on PR. PS = 3 cm, SQ = 6 cm and ST = 4 cm. Find QR.
Given: PS/PQ = 3/9 = 1/3
Use: ΔPST ~ ΔPQR (AA)
ST/QR = PS/PQ = 1/3
QR = 3 × 4 = 12
Answer: QR = 12 cm
D and E are the mid-points of AB and AC of ΔABC, and BC = 10 cm. Show DE ∥ BC and find DE.
Given: AD/DB = AE/EC = 1
By the converse of BPT, DE ∥ BC.
Then ΔADE ~ ΔABC (AA), so DE/BC = AD/AB = 1/2.
Answer: DE ∥ BC and DE = 5 cm
ΔABC ~ ΔPQR with AB = 4 cm, BC = 6 cm, CA = 5 cm and PQ = 6 cm. Find QR and RP.
Given: scale factor PQ/AB = 6/4 = 3/2
Use: Corresponding sides are proportional
QR = 3/2 × 6 = 9
RP = 3/2 × 5 = 7.5
Answer: QR = 9 cm, RP = 7.5 cm
Triangle X has sides 3 cm, 4 cm, 5 cm and triangle Y has sides 6 cm, 8 cm, 10 cm. Are they similar? By which criterion?
Use: Compare the ratios of sides in order
3/6 = 4/8 = 5/10 = 1/2
Answer: Yes, by SSS
In ΔABC, ∠A = 50° and ∠B = 60°. In ΔDEF, ∠D = 50° and ∠F = 70°. Are the triangles similar? Write the correspondence.
Use: Angle sum 180°
∠C = 70° and ∠E = 60°.
So ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
Answer: Yes, ΔABC ~ ΔDEF (AAA)
ΔABC ~ ΔDEF. Their perimeters are 30 cm and 18 cm, and AB = 10 cm. Find DE.
Use: Perimeters of similar triangles are in the ratio of corresponding sides
If every side of DEF is k times the matching side of ABC, the perimeter is k times too: k = 18/30 = 3/5.
DE = 3/5 × 10 = 6
Answer: DE = 6 cm
In ΔABC, DE ∥ BC with AD : DB = 2 : 3. Find DE : BC.
Given: AD/AB = 2/5
Use: ΔADE ~ ΔABC (AA: common ∠A, corresponding angles)
DE/BC = AD/AB = 2/5
Answer: DE : BC = 2 : 5
ABCD is a trapezium with AB ∥ DC and AB = 2 CD. Its diagonals meet at O and AO = 6 cm. Find OC.
Use: ΔAOB ~ ΔCOD
∠OAB = ∠OCD and ∠OBA = ∠ODC (alternate angles), so ΔAOB ~ ΔCOD (AA).
AO/OC = AB/CD = 2 ⇒ OC = 3
Answer: OC = 3 cm
A girl 1.5 m tall casts a shadow 2 m long. At the same time a tree casts a shadow 24 m long. Find the height of the tree.
Use: Similar right triangles (AA)
h/24 = 1.5/2 ⇒ h = 18
Answer: 18 m
A girl 90 cm tall walks away from the base of a 3.6 m lamp-post at 1.2 m/s. Find the length of her shadow after 4 seconds.
Given: distance walked = 1.2 × 4 = 4.8 m
Use: The girl and the lamp-post with their shadow tips form similar triangles
x/(x + 4.8) = 0.9/3.6 = 1/4
4x = x + 4.8 ⇒ x = 1.6
Answer: 1.6 m
In ΔABC, D is on BC with ∠ADC = ∠BAC. If CB = 8 cm and CD = 2 cm, find CA.
Use: ΔADC ~ ΔBAC (∠C common, ∠ADC = ∠BAC)
CA/CB = CD/CA ⇒ CA² = CB · CD = 16
Answer: CA = 4 cm
In ΔABC, AD ⊥ BC and CE ⊥ AB. AB = 10 cm, AD = 6 cm and CE = 8 cm. Find BC.
Use: ΔABD ~ ΔCBE (∠B common, right angles at D and E)
AB/CB = AD/CE ⇒ 10/BC = 6/8
BC = 80/6 = 40/3
Answer: BC = 40/3 cm ≈ 13.3 cm
D and E lie on AB and AC of ΔABC with AD × EC = AE × DB. Prove that DE ∥ BC.
Given: AD · EC = AE · DB
Use: Converse of BPT
Dividing both sides by DB · EC: AD/DB = AE/EC.
So DE divides AB and AC in the same ratio, and DE ∥ BC (Theorem 6.2).
Answer: Proved
In ΔABC, ∠B = 90° and BD ⊥ AC. Prove that ΔADB ~ ΔABC and AB² = AD · AC.
Use: AA
In ΔADB and ΔABC: ∠A is common and ∠ADB = ∠ABC = 90°.
So ΔADB ~ ΔABC, giving AD/AB = AB/AC.
Hence AB² = AD · AC.
Answer: Proved
P and Q are points on sides AB and AC of ΔABC with AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm. Show that ΔAPQ ~ ΔABC and find PQ if BC = 9 cm.
Given: AP/AB = 2/6 = 1/3, AQ/AC = 3/9 = 1/3
Use: SAS
∠A is common and the sides including it are in the same ratio 1/3, so ΔAPQ ~ ΔABC.
PQ/BC = 1/3 ⇒ PQ = 3
Answer: PQ = 3 cm
In ΔABC, X is on AB and Y on AC with XY ∥ BC and AX/XB = 1/2. Find AY/YC and XY/BC.
Given: AX : XB = 1 : 2
Use: BPT and AA
AY/YC = AX/XB = 1/2 (BPT)
XY/BC = AX/AB = 1/3 (ΔAXY ~ ΔABC)
Answer: AY/YC = 1/2; XY/BC = 1/3
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Answer: (i) similar (ii) similar (iii) equilateral (iv) (a) equal (b) proportional
(i) Similar: two circles of radii 2 cm and 5 cm; two equilateral triangles of sides 3 cm and 6 cm; a photograph and its enlargement.
(ii) Non-similar: a square and a rectangle that is not a square; a square and a rhombus that is not a square; an equilateral triangle and a right triangle.
Answer: as above (other examples are also correct)
Sides: PQ/AB = QR/BC = RS/CD = SP/DA = 1.5/3 = 1/2, so the sides are proportional.
Angles: all angles of ABCD are 90°, but the angles of the rhombus PQRS are not 90°. So the corresponding angles are not equal.
Answer: Not similar
(i) By BPT, AD/DB = AE/EC: 1.5/3 = 1/EC ⇒ EC = 2 cm.
(ii) AD/DB = AE/EC: AD/7.2 = 1.8/5.4 ⇒ AD = 7.2 × 1/3 = 2.4 cm.
Answer: (i) EC = 2 cm (ii) AD = 2.4 cm
(i) PE/EQ = 3.9/3 = 1.3 and PF/FR = 3.6/2.4 = 1.5. Not equal, so EF is not parallel to QR.
(ii) PE/EQ = 4/4.5 = 8/9 and PF/FR = 8/9. Equal, so by the converse of BPT, EF ∥ QR.
(iii) EQ = 1.28 − 0.18 = 1.10 and FR = 2.56 − 0.36 = 2.20. PE/EQ = 0.18/1.10 = 9/55 and PF/FR = 0.36/2.20 = 9/55. Equal, so EF ∥ QR.
Answer: (i) not parallel (ii) parallel (iii) parallel
In ΔABC, LM ∥ CB, so by BPT AMMB = ALLC, which gives AMAB = ALAC ... (1)
In ΔADC, LN ∥ CD, so ANND = ALLC, which gives ANAD = ALAC ... (2)
From (1) and (2), AMAB = ANAD.
Answer: Proved
In ΔABC, DE ∥ AC, so BDDA = BEEC ... (1) (BPT)
In ΔABE, DF ∥ AE, so BDDA = BFFE ... (2) (BPT)
From (1) and (2), BFFE = BEEC.
Answer: Proved
In ΔPOQ, DE ∥ OQ, so PEEQ = PDDO ... (1)
In ΔPOR, DF ∥ OR, so PFFR = PDDO ... (2)
From (1) and (2), PEEQ = PFFR. So in ΔPQR, EF divides PQ and PR in the same ratio, and by Theorem 6.2, EF ∥ QR.
Answer: Proved
In ΔOPQ, AB ∥ PQ, so OAAP = OBBQ ... (1)
In ΔOPR, AC ∥ PR, so OAAP = OCCR ... (2)
From (1) and (2), OBBQ = OCCR. So in ΔOQR, BC ∥ QR by Theorem 6.2.
Answer: Proved
Let D be the mid-point of AB in ΔABC, and let the line through D parallel to BC meet AC at E.
By Theorem 6.1, ADDB = AEEC.
Since AD = DB, the left side is 1, so AE = EC. Hence E is the mid-point of AC.
Answer: Proved
Let D and E be the mid-points of AB and AC of ΔABC. Then AD = DB and AE = EC.
So ADDB = 1 = AEEC. DE divides AB and AC in the same ratio, so by Theorem 6.2, DE ∥ BC.
Answer: Proved
Draw EO ∥ DC (and so ∥ AB) through O, meeting AD at E.
In ΔADC, EO ∥ DC, so AEED = AOOC ... (1)
In ΔDAB, EO ∥ AB, so DEEA = DOOB, i.e. AEED = BOOD ... (2)
From (1) and (2), AOOC = BOOD, so AOBO = CODO.
Answer: Proved
Given AOBO = CODO, so AOOC = BOOD ... (1)
Draw OE ∥ AB through O, meeting AD at E.
In ΔABD, OE ∥ AB, so AEED = BOOD ... (2)
From (1) and (2), AEED = AOOC. So in ΔADC, EO divides AD and AC in the same ratio, and EO ∥ DC (Theorem 6.2).
Thus AB ∥ EO ∥ DC, so AB ∥ DC and ABCD is a trapezium.
Answer: Proved
(i) ∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°: ΔABC ~ ΔPQR (AAA).
(ii) AB/QR = 2/4, BC/RP = 2.5/5, CA/PQ = 3/6, all equal to 1/2: ΔABC ~ ΔQRP (SSS).
(iii) Sides 2, 2.7, 3 and 4, 5, 6: 2/4 = 0.5 but 2.7/5 = 0.54. Not similar.
(iv) ∠M = ∠Q = 70° and MN/QP = 2.5/5 = 1/2, ML/QR = 5/10 = 1/2: ΔNML ~ ΔPQR (SAS).
(v) The 80° angle at A is not included between the given sides AB and BC, so SAS cannot be applied: not similar (cannot be concluded).
(vi) In ΔDEF, ∠F = 180° − 70° − 80° = 30°. In ΔPQR, ∠P = 180° − 80° − 30° = 70°. So ∠D = ∠P, ∠E = ∠Q, ∠F = ∠R: ΔDEF ~ ΔPQR (AAA).
Answer: (i) ΔABC ~ ΔPQR (AAA) (ii) ΔABC ~ ΔQRP (SSS) (iii) not similar (iv) ΔNML ~ ΔPQR (SAS) (v) not similar (vi) ΔDEF ~ ΔPQR (AAA)
DOB is a straight line, so ∠DOC = 180° − 125° = 55°.
In ΔODC, ∠DCO = 180° − 70° − 55° = 55°.
ΔODC ~ ΔOBA gives ∠OAB = ∠OCD = 55°.
Answer: ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°
In ΔAOB and ΔCOD:
∠OAB = ∠OCD (alternate angles, AB ∥ DC)
∠OBA = ∠ODC (alternate angles)
So ΔAOB ~ ΔCOD (AA), and OAOC = OBOD.
Answer: Proved
In ΔPQR, ∠PQR = ∠PRQ (∠1 = ∠2), so PQ = PR (sides opposite equal angles).
Then QRQS = QTPQ, i.e. QSQR = QPQT.
In ΔPQS and ΔTQR: ∠Q is common and the sides including it are proportional.
So ΔPQS ~ ΔTQR (SAS).
Answer: Proved
In ΔRPQ and ΔRTS:
∠RPQ = ∠RTS (given)
∠PRQ = ∠TRS (common angle R)
So ΔRPQ ~ ΔRTS (AA).
Answer: Proved
ΔABE ≅ ΔACD gives AB = AC and AE = AD (CPCT).
So ADAB = AEAC.
In ΔADE and ΔABC, ∠A is common and the including sides are proportional.
So ΔADE ~ ΔABC (SAS).
Answer: Proved
(i) ∠AEP = ∠CDP = 90° and ∠APE = ∠CPD (vertically opposite). So ΔAEP ~ ΔCDP (AA).
(ii) ∠ADB = ∠CEB = 90° and ∠ABD = ∠CBE (common). So ΔABD ~ ΔCBE (AA).
(iii) ∠AEP = ∠ADB = 90° and ∠PAE = ∠BAD (common). So ΔAEP ~ ΔADB (AA).
(iv) ∠PDC = ∠BEC = 90° and ∠PCD = ∠BCE (common). So ΔPDC ~ ΔBEC (AA).
Answer: Proved
In ΔABE and ΔCFB:
∠A = ∠C (opposite angles of a parallelogram)
∠AEB = ∠CBF (alternate angles, since AE ∥ BC)
So ΔABE ~ ΔCFB (AA).
Answer: Proved
(i) ∠ABC = ∠AMP = 90° and ∠BAC = ∠MAP (common). So ΔABC ~ ΔAMP (AA).
(ii) Corresponding sides of similar triangles are proportional: CAPA = BCMP.
Answer: Proved
ΔABC ~ ΔFEG gives ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE. Halving the last: ∠ACD = ∠FGH and ∠DCB = ∠HGE.
(ii) In ΔDCB and ΔHGE: ∠DCB = ∠HGE and ∠B = ∠E. So ΔDCB ~ ΔHGE (AA).
(iii) In ΔDCA and ΔHGF: ∠ACD = ∠FGH and ∠A = ∠F. So ΔDCA ~ ΔHGF (AA).
(i) From (iii), CDGH = ACFG.
Answer: Proved
AB = AC, so ∠ABC = ∠ACB. Hence ∠ABD = ∠ECF.
∠ADB = ∠EFC = 90°.
So ΔABD ~ ΔECF (AA).
Answer: Proved
Given ABPQ = BCQR = ADPM. Since BD = ½BC and QM = ½QR, BCQR = BDQM.
So ABPQ = BDQM = ADPM, and ΔABD ~ ΔPQM (SSS). Hence ∠B = ∠Q.
In ΔABC and ΔPQR: ∠B = ∠Q and ABPQ = BCQR. So ΔABC ~ ΔPQR (SAS).
Answer: Proved
In ΔADC and ΔBAC: ∠ADC = ∠BAC (given) and ∠ACD = ∠BCA (common).
So ΔADC ~ ΔBAC (AA).
Hence CACB = CDCA, so CA² = CB · CD.
Answer: Proved
Produce AD to E with DE = AD and join CE; produce PM to N with MN = PM and join RN.
ΔABD ≅ ΔECD (BD = DC, AD = DE, vertically opposite angles), so AB = CE and ∠BAD = ∠CED. Similarly PQ = RN and ∠QPM = ∠RNM.
Now ACPR = ABPQ = CERN and AEPN = 2AD2PM = ADPM = ABPQ.
So ΔACE ~ ΔPRN (SSS), giving ∠CAE = ∠RPN and ∠CEA = ∠RNP.
Then ∠BAD = ∠CEA = ∠RNP = ∠QPM. Adding, ∠BAC = ∠BAD + ∠CAD = ∠QPM + ∠RPM = ∠QPR.
In ΔABC and ΔPQR: ∠A = ∠P and ABPQ = ACPR. So ΔABC ~ ΔPQR (SAS).
Answer: Proved
At the same time the sun's rays make equal angles with the ground, and both the pole and the tower are vertical. So the two right triangles are similar (AA).
h6 = 284 ⇒ h = 42 m.
Answer: 42 m
ΔABC ~ ΔPQR gives ∠B = ∠Q and ABPQ = BCQR = 2BD2QM = BDQM.
In ΔABD and ΔPQM: ∠B = ∠Q and the including sides are proportional, so ΔABD ~ ΔPQM (SAS).
Hence ABPQ = ADPM.
Answer: Proved