Important Notes: Triangles
🔑 Key Definitions You Must Remember
- Similar Figures: Two figures that have the exact same shape but can have different sizes. Think of two photos of the same object — one small, one enlarged.
- Congruent vs Similar: All congruent figures are automatically similar (same shape AND size), but similar figures need not be congruent (sizes can differ).
- Polygon Similarity Rule: Two polygons with the same number of sides are similar ONLY IF:
- (i) Their corresponding angles are equal
- (ii) Their corresponding sides are proportional (in the same ratio)
📐 Basic Proportionality Theorem (BPT) — Also called Thales Theorem
If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then it divides those two sides in the same ratio.
If DE || BC in ΔABC, then AD/DB = AE/EC
🔄 Converse of BPT
If a line divides any two sides of a triangle in the same ratio, then that line must be parallel to the third side.
If AD/DB = AE/EC, then DE || BC
📊 Similarity Criteria for Triangles
- AAA (Angle-Angle-Angle): If all three corresponding angles of two triangles are equal, then their corresponding sides are automatically in the same ratio → triangles are similar.
- AA (Angle-Angle): If just two angles of one triangle equal two angles of another triangle, the third angles must also be equal (since angles sum to 180°). So AA is enough!
- SSS Similarity: If all three corresponding sides of two triangles are in the same ratio, the triangles are similar.
- SAS Similarity: If one angle is equal and the sides including that angle are proportional, the triangles are similar.
📏 Area Theorem for Similar Triangles
The ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides.
ar(ΔABC) / ar(ΔPQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²
📐 Pythagoras Theorem
In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.
AC² = AB² + BC² (where ∠B = 90°)
Exercise 6.1 — Introduction to Similarity
1
Fill in the blanks using the correct word given in brackets:
(i) All circles are _______. (congruent, similar)
✅ Correct Answer: Similar
Understand the shape: Every circle has the exact same shape — perfectly round with 360° at the center. No matter how big or small, a circle always looks the same.
Check congruence: Two circles are congruent only if they have the same radius. A circle with radius 2 cm and a circle with radius 5 cm are clearly different in size, so they are NOT congruent.
Check similarity: For similarity, we only need the same shape. Since all circles have the same shape (only size varies), they satisfy the definition of similar figures.
Conclusion: All circles are similar because they have the same shape but may have different sizes.
(ii) All squares are _______. (similar, congruent)
✅ Correct Answer: Similar
Analyze the angles: Every square has four angles, and each angle is exactly 90°. So corresponding angles of any two squares are always equal.
Analyze the sides: In a square, all four sides are equal. If one square has side 2 cm and another has side 5 cm, the ratio of corresponding sides is 2/5 for every pair. This ratio is constant.
Why not congruent? A square of side 2 cm and a square of side 3 cm have different sizes. Congruent figures must be identical in both shape AND size. So they are not congruent.
Conclusion: Since angles are equal and sides are proportional, all squares are similar.
(iii) All _______ triangles are similar. (isosceles, equilateral)
✅ Correct Answer: Equilateral
Understand equilateral triangles: In an equilateral triangle, all three sides are equal AND all three angles are equal to 60°.
Check similarity of two equilateral triangles: Take any two equilateral triangles — Triangle 1 with side 3 cm and Triangle 2 with side 7 cm. All angles are 60° in both, so corresponding angles are equal. The ratio of sides is 3/7 for every pair. By AAA criterion, they are similar.
Why not isosceles? An isosceles triangle has only two equal sides. One isosceles triangle can have angles 70°, 70°, 40° while another can have 50°, 50°, 80°. Their corresponding angles are NOT equal, so they are NOT necessarily similar.
Conclusion: All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _______ and (b) their corresponding sides are _______. (equal, proportional)
✅ Correct Answer: Equal, Proportional
Condition (a) — Angles: For two polygons to have the same shape, their corresponding angles must be exactly equal. If angles differ, the shape changes (e.g., a square vs a rhombus — both have 4 equal sides but angles differ, so shapes differ).
Condition (b) — Sides: The corresponding sides must be in the same ratio (proportional). This means if one polygon is an enlarged or reduced version of the other, the scaling factor must be consistent for all sides.
Why both are needed: Consider a square (all angles 90°, sides equal) and a rectangle (all angles 90°, but sides not all equal). Angles are equal but sides are not proportional — so they are NOT similar. Similarly, a square and a rhombus have proportional sides but different angles — NOT similar.
Conclusion: (a) Equal and (b) Proportional.
2
Give two different examples of pair of:
(i) similar figures (ii) non-similar figures
✅ Correct Answer: B
Part (i) — Similar Figures: We need pairs that have the same shape but can have different sizes.
• Two equilateral triangles: All angles are 60° in both, and sides are in constant ratio. By AAA, they are similar.
• Two squares: All angles are 90° in both, and sides are in constant ratio. They are similar.
• Two equilateral triangles: All angles are 60° in both, and sides are in constant ratio. By AAA, they are similar.
• Two squares: All angles are 90° in both, and sides are in constant ratio. They are similar.
Part (ii) — Non-Similar Figures: We need pairs that do NOT have the same shape.
• Square and Rectangle: Both have 90° angles, but in a square all sides are equal while in a rectangle opposite sides are equal (adjacent sides may differ). The sides are NOT in the same ratio for all pairs. So they are NOT similar.
• Circle and Ellipse: A circle has constant radius in all directions. An ellipse is stretched in one direction. They have completely different shapes. NOT similar.
• Square and Rectangle: Both have 90° angles, but in a square all sides are equal while in a rectangle opposite sides are equal (adjacent sides may differ). The sides are NOT in the same ratio for all pairs. So they are NOT similar.
• Circle and Ellipse: A circle has constant radius in all directions. An ellipse is stretched in one direction. They have completely different shapes. NOT similar.
3
State whether the following quadrilaterals are similar or not (Fig. 6.8):
✅ Correct Answer: Not Similar
Identify the figures: On the left is a rhombus PQRS with all sides = 1.5 cm. On the right is a square ABCD with all sides = 3 cm.
Check side ratios: The ratio of corresponding sides is 1.5/3 = 1/2 for all sides. So the sides ARE proportional. This satisfies one condition.
Check angles: In square ABCD, all four angles are exactly 90°. In rhombus PQRS, the angles are NOT 90° (a rhombus has opposite angles equal, but adjacent angles are supplementary — they add to 180° but are not 90° unless it's a square).
Apply the similarity rule: For two polygons to be similar, BOTH conditions must be satisfied: (a) corresponding angles equal AND (b) corresponding sides proportional. Here, condition (a) fails.
Conclusion: Since corresponding angles are not equal, the quadrilaterals are NOT similar. This proves that having proportional sides alone is NOT enough for similarity.
Exercise 6.2 — Basic Proportionality Theorem (BPT)
2
E and F are points on sides PQ and PR of ΔPQR. For each case, state whether EF || QR:
Remember the Converse of BPT: If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
✅ Correct Answer: EF is NOT parallel to QR
Given
In ΔPQR, E is a point on PQ and F is a point on PR such that:
- PE = 3.9 cm
- EQ = 3 cm
- PF = 3.6 cm
- FR = 2.4 cm
To Check
Whether EF || QR using the Converse of Basic Proportionality Theorem.
Proof
Calculate the ratio PE/EQ:
PE/EQ = 3.9/3 = 1.3
Calculate the ratio PF/FR:
PF/FR = 3.6/2.4 = 1.5
Compare the two ratios:
PE/EQ = 1.3
and PF/FR = 1.5
Since 1.3 ≠ 1.5, the two sides are NOT divided in the same ratio.
Conclusion: Since PE/EQ ≠ PF/FR, by the Converse of BPT, EF is NOT parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
✅ Correct Answer: EF || QR
Given
In ΔPQR, E is on PQ and F is on PR such that:
- PE = 4 cm, QE = 4.5 cm
- PF = 8 cm, RF = 9 cm
To Check
Whether EF || QR using Converse of BPT.
Proof
Calculate PE/QE:
PE/QE = 4/4.5 = 4/(9/2) = 8/9
[Multiplying numerator and denominator by 2]
Calculate PF/RF:
PF/RF = 8/9
Compare:
PE/QE = PF/RF = 8/9
Both ratios are exactly equal.
Conclusion: Since PE/QE = PF/RF, by the Converse of BPT, EF || QR. Hence proved.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
✅ Correct Answer: EF || QR
Given
In ΔPQR:
- PQ = 1.28 cm, PR = 2.56 cm
- PE = 0.18 cm, PF = 0.36 cm
To Check
Whether EF || QR. We need EQ and FR first.
Proof
Find EQ using PQ = PE + EQ:
EQ = PQ − PE = 1.28 − 0.18 = 1.10 cm
Find FR using PR = PF + FR:
FR = PR − PF = 2.56 − 0.36 = 2.20 cm
Calculate PE/EQ:
PE/EQ = 0.18/1.10 = 18/110 = 9/55
Calculate PF/FR:
PF/FR = 0.36/2.20 = 36/220 = 9/55
Compare:
PE/EQ = PF/FR = 9/55
Both ratios are exactly equal.
Conclusion: Since PE/EQ = PF/FR, by Converse of BPT, EF || QR. Hence proved.
3
In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Use BPT in two different triangles that share the common ratio AL/AC.
✅ Correct Approach: Apply BPT in ΔABC and ΔACD
Given
In the figure, LM || CB and LN || CD, where M is on AB, L is on AC, and N is on AD.
To Prove
AM/AB = AN/AD
Proof
In ΔABC, since LM || CB (Given), by Basic Proportionality Theorem:
AM/AB = AL/AC
[BPT: A line parallel to one side divides the other two sides proportionally]
Let us call this Equation (1).
In ΔACD, since LN || CD (Given), by Basic Proportionality Theorem:
AN/AD = AL/AC
[Same reason as above]
Let us call this Equation (2).
From Equation (1) and Equation (2):
AM/AB = AL/AC = AN/AD
[Both equal AL/AC, so they equal each other]
Therefore, AM/AB = AN/AD. Hence Proved.
4
In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
You need to use BPT twice — once in a triangle containing DE, and once in a triangle containing DF. Both should give you BD/DA.
✅ Correct Approach: Apply BPT in ΔBAE and ΔBCA
Given
In ΔBAC:
- DE || AC (Given)
- DF || AE (Given)
Where D is on BA, F is on BE, and E is on BC.
To Prove
BF/FE = BE/EC
Proof
In ΔBAE, since DF || AE (Given), by Basic Proportionality Theorem:
BF/FE = BD/DA
[Line parallel to one side divides other two sides proportionally]
Call this Equation (1).
In ΔBCA, since DE || AC (Given), by Basic Proportionality Theorem:
BE/EC = BD/DA
[Same theorem applied to larger triangle]
Call this Equation (2).
From Equation (1) and Equation (2):
BF/FE = BD/DA = BE/EC
[Both left sides equal BD/DA]
Therefore, BF/FE = BE/EC. Hence Proved.
5
In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
✅ Correct Approach: Apply BPT in ΔPOQ and ΔPOR
Given
In ΔPQR, with point O inside:
- DE || OQ (Given)
- DF || OR (Given)
Where D is on PO, E is on PQ, and F is on PR.
To Prove
EF || QR
Proof
In ΔPOQ, since DE || OQ (Given), by BPT:
PE/EQ = PD/DO
[Line parallel to base divides sides proportionally]
Call this Equation (1).
In ΔPOR, since DF || OR (Given), by BPT:
PF/FR = PD/DO
[Same theorem applied to second triangle]
Call this Equation (2).
From Equation (1) and Equation (2):
PE/EQ = PF/FR
[Both equal PD/DO]
Now consider ΔPQR. The line EF intersects PQ at E and PR at F.
We have shown that EF divides sides PQ and PR in the same ratio.
By the Converse of Basic Proportionality Theorem:
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Therefore, EF || QR. Hence Proved.
7
Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Theorem 6.1 is the Basic Proportionality Theorem (BPT). If D is the mid-point, then AD = DB, so AD/DB = 1.
✅ Correct Answer: A
Given
ΔABC in which D is the mid-point of side AB.
A line DE is drawn through D such that DE || BC, with E on AC.
To Prove
E is the mid-point of AC, i.e.,
AE = EC
or AE/EC = 1
Proof
Since D is the mid-point of AB (Given):
AD = DB
[Definition of mid-point]
Therefore:
AD/DB = 1
[Dividing equal quantities]
In ΔABC, since DE || BC (Given), by Theorem 6.1 (BPT):
AD/DB = AE/EC
[BPT: Parallel line divides sides proportionally]
From steps 2 and 3:
AE/EC = 1
[Since AD/DB = 1 and AD/DB = AE/EC]
Therefore:
AE = EC
[If ratio is 1, the quantities are equal]
Therefore, E is the mid-point of AC. Hence Proved.
9
ABCD is a trapezium with AB || DC. Diagonals intersect at O. Show that AO/BO = CO/DO.
Look for two triangles formed by the diagonals that are similar. Use the fact that AB || DC creates alternate interior angles.
✅ Correct Answer: Prove ΔAOB ~ ΔCOD by AA
Given
ABCD is a trapezium with AB || DC.
Diagonals AC and BD intersect at point O.
To Prove
AO/BO = CO/DO
Proof
In ΔAOB and ΔCOD:
∠AOB = ∠COD
[Vertically opposite angles are always equal]Since AB || DC (Given) and AC is a transversal:
∠OAB = ∠OCD
[Alternate interior angles are equal when lines are parallel]Since AB || DC (Given) and BD is a transversal:
∠OBA = ∠ODC
[Alternate interior angles are equal when lines are parallel]By AAA similarity criterion:
ΔAOB ~ ΔCOD
[All three corresponding angles are equal]
Since the triangles are similar, corresponding sides are proportional:
AO/CO = BO/DO = AB/CD
[Property of similar triangles]
From AO/CO = BO/DO, cross-multiplying:
AO × DO = BO × CO
Dividing both sides by BO × DO:
AO/BO = CO/DO
Therefore, AO/BO = CO/DO. Hence Proved.
Exercise 6.3 — Criteria for Similarity of Triangles
1
State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion and symbolic form.
(i) ΔABC and ΔPQR with angles shown
✅ Correct Answer: ΔABC ~ ΔPQR by AAA
Given
In ΔABC: ∠A = 60°, ∠B = 80°, ∠C = 40°
In ΔPQR: ∠P = 60°, ∠Q = 80°, ∠R = 40°
Proof
Compare corresponding angles:
∠A = 60° = ∠P
∠B = 80° = ∠Q
∠C = 40° = ∠R
Since all three corresponding angles are equal, by AAA similarity criterion:
ΔABC ~ ΔPQR
(ii) ΔABC and ΔPQR with sides shown
✅ Correct Answer: ΔABC ~ ΔQPR by SSS
Given
ΔABC: AB = 2, BC = 2.5, AC = 3
ΔPQR: PQ = 4, QR = 6, PR = 5
Proof
Match shortest sides: AB = 2 and PQ = 4
AB/QP = 2/4 = 1/2
Match longest sides: AC = 3 and QR = 6
AC/QR = 3/6 = 1/2
Match remaining sides: BC = 2.5 and PR = 5
BC/PR = 2.5/5 = 1/2
Since all three ratios equal 1/2, by SSS similarity criterion:
ΔABC ~ ΔQPR
[A↔Q, B↔P, C↔R]
(iii) ΔLMP and ΔDEF
✅ Correct Answer: Not Similar
Given
ΔLMP: LM = 2.7, MP = 2, LP = 3
ΔDEF: DE = 5, EF = 4, DF = 6
Proof
Calculate ratios of corresponding sides:
LM/DE = 2.7/5 = 0.54
MP/EF = 2/4 = 0.50
LP/DF = 3/6 = 0.50
Since 0.54 ≠ 0.50, the ratios are NOT all equal.
For SSS similarity, all three ratios must be equal.
Therefore, ΔLMP is NOT similar to ΔDEF.
(iv) ΔMNL and ΔPQR
✅ Correct Answer: ΔMNL ~ ΔPQR by SAS
Given
ΔMNL: MN = 2.5, ML = 5, ∠M = 70°
ΔPQR: PQ = 5, PR = 10, ∠P = 70°
Proof
Compare the included angles:
∠M = ∠P = 70°
Compare the sides including this angle:
MN/PQ = 2.5/5 = 1/2
ML/PR = 5/10 = 1/2
Since ∠M = ∠P and the including sides are proportional, by SAS similarity criterion:
ΔMNL ~ ΔPQR
(v) ΔABC and ΔDEF
✅ Correct Answer: ΔABC ~ ΔDFE by SAS
Given
ΔABC: AB = 2.5, BC = 3, ∠B = 80°
ΔDEF: DE = 6, EF = 5, ∠F = 80°
Proof
Identify corresponding angles: ∠B = 80° and ∠F = 80°, so B ↔ F.
Check sides including these angles:
AB/DF = 2.5/5 = 1/2
BC/FE = 3/6 = 1/2
Since ∠B = ∠F and including sides are proportional, by SAS:
ΔABC ~ ΔDFE
(vi) ΔDEF and ΔPQR
✅ Correct Answer: ΔDEF ~ ΔPQR by AAA
Given
ΔDEF: ∠D = 70°, ∠E = 80°
ΔPQR: ∠Q = 80°, ∠R = 30°
Proof
Find third angle of ΔDEF:
∠F = 180° − 70° − 80° = 30°
Find third angle of ΔPQR:
∠P = 180° − 80° − 30° = 70°
Compare all angles:
∠D = 70° = ∠P
∠E = 80° = ∠Q
∠F = 30° = ∠R
By AAA similarity criterion:
ΔDEF ~ ΔPQR
2
In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125°, ∠CDO = 70°. Find ∠DOC, ∠DCO, and ∠OAB.
✅ Correct Answer: 55°, 55°, 55°
Given
ΔODC ~ ΔOBA, ∠BOC = 125°, ∠CDO = 70°
Solution
Find ∠DOC:
∠DOC + ∠BOC = 180°
[Linear pair]
∠DOC = 180° − 125° = 55°
Find ∠DCO using angle sum in ΔODC:
∠DCO = 180° − ∠CDO − ∠DOC
∠DCO = 180° − 70° − 55° = 55°
Find ∠OAB using similarity:
Since ΔODC ~ ΔOBA, corresponding angles are equal.
∠OAB = ∠OCD = ∠DCO = 55°
∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°
3
Diagonals AC and BD of trapezium ABCD (AB || DC) intersect at O. Using similarity criterion, show that OA/OC = OB/OD.
✅ Correct Answer: ΔAOB ~ ΔCOD by AA
Given
ABCD is a trapezium with AB || DC.
Diagonals AC and BD intersect at O.
To Prove
OA/OC = OB/OD
Proof
In ΔAOB and ΔCOD:
∠AOB = ∠COD
[Vertically opposite angles]Since AB || DC and AC is transversal:
∠OAB = ∠OCD
[Alternate interior angles]By AA similarity:
ΔAOB ~ ΔCOD
Therefore, corresponding sides are proportional:
OA/OC = OB/OD = AB/CD
Hence, OA/OC = OB/OD. Proved.
5
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
✅ Correct Answer: By AA
Given
S is on PR and T is on QR of ΔPQR.
∠P = ∠RTS
To Prove
ΔRPQ ~ ΔRTS
Proof
In ΔRPQ and ΔRTS:
∠PRQ = ∠TRS
[Common angle R]∠RPQ = ∠RTS
[Given]
By AA similarity criterion:
ΔRPQ ~ ΔRTS
Hence Proved.
6
In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
✅ Correct Answer: By SAS
Given
ΔABE ≅ ΔACD
To Prove
ΔADE ~ ΔABC
Proof
Since ΔABE ≅ ΔACD, by CPCT:
AB = AC
and AE = AD
Therefore:
AD/AB = AE/AC
[Since AD = AE and AB = AC]
Also, ∠DAE = ∠BAC
[Common angle A]
By SAS similarity criterion:
ΔADE ~ ΔABC
Hence Proved.
9
ABC and AMP are right triangles, right-angled at B and M. Prove that (i) ΔABC ~ ΔAMP and (ii) CA/PA = BC/MP.
✅ Correct Answer: AA
Given
ΔABC is right-angled at B.
ΔAMP is right-angled at M.
To Prove
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Proof
Part (i):
In ΔABC and ΔAMP:
∠ABC = ∠AMP = 90°
[Given]∠BAC = ∠MAP
[Common angle A]
By AA similarity criterion:
ΔABC ~ ΔAMP
Part (ii): Since ΔABC ~ ΔAMP, corresponding sides are proportional:
AB/AM = BC/MP = CA/PA
From this:
CA/PA = BC/MP
Hence Proved.
13
D is a point on side BC of ΔABC such that ∠ADC = ∠BAC. Show that CA² = CB·CD.
✅ Correct Answer: Prove ΔADC ~ ΔBAC by AA
Given
D is on BC of ΔABC such that ∠ADC = ∠BAC.
To Prove
CA² = CB · CD
Proof
In ΔADC and ΔBAC:
∠ACD = ∠BCA
[Common angle C]∠ADC = ∠BAC
[Given]
By AA similarity criterion:
ΔADC ~ ΔBAC
Therefore, corresponding sides are proportional:
AD/BA = DC/AC = AC/BC
From DC/AC = AC/BC:
AC² = BC × DC
[Cross-multiplication]
Or:
CA² = CB × CD
Hence Proved.
15
A vertical pole of length 6 m casts a shadow 4 m long. At the same time, a tower casts a shadow 28 m long. Find the height of the tower.
✅ Correct Answer: 42 m
Given
Pole: Height = 6 m, Shadow = 4 m
Tower: Shadow = 28 m, Height = h m
Solution
At the same time, the sun's angle of elevation is the same for both objects. The triangles formed by the objects and their shadows are similar by AA criterion (right angle + common sun angle).
For similar triangles, height-to-shadow ratio is constant:
6/4 = h/28
Solve for h:
h = (6 × 28)/4 = 168/4 = 42
Height of tower = 42 meters
16
If AD and PM are medians of ΔABC and ΔPQR, and ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM.
✅ Correct Answer: Show ΔABD ~ ΔPQM by SAS
Given
ΔABC ~ ΔPQR
AD is median of ΔABC, PM is median of ΔPQR.
To Prove
AB/PQ = AD/PM
Proof
Since ΔABC ~ ΔPQR:
AB/PQ = BC/QR = AC/PR = k
[Corresponding sides of similar triangles]
Since AD and PM are medians:
BD = BC/2
and QM = QR/2
Therefore:
BD/QM = (BC/2)/(QR/2) = BC/QR = AB/PQ
Also, ∠B = ∠Q
[Corresponding angles of similar triangles]
In ΔABD and ΔPQM:
AB/PQ = BD/QM (proved)
∠B = ∠Q (proved)
By SAS similarity, ΔABD ~ ΔPQMTherefore:
AB/PQ = AD/PM
Hence Proved.
🧮 Interactive Calculators
📐 BPT Parallel Line Checker
Check if EF is parallel to QR using the Converse of Basic Proportionality Theorem. If PE/EQ = PF/FR, then EF || QR.
📊 SSS Similarity Checker
Enter all three sides of two triangles to check if they are similar by SSS criterion. The ratios of corresponding sides must all be equal.
☀️ Height & Shadow Calculator
Solve real-world problems like Exercise 6.3 Q15. If two objects cast shadows at the same time, their height-to-shadow ratios are equal (similar triangles).
🔺 AA Similarity Angle Checker
Enter two angles of each triangle. If two angles of one triangle equal two angles of another, the triangles are similar by AA criterion (the third angle is automatically equal since angles sum to 180°).