Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers correspondence, similarity criteria, BPT applications and indirect measurement. Complete deductive proofs of BPT and all similarity criteria, and every textbook problem, are still outside this companion.
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LESSON 1
Match corresponding parts
Use AA, SAS and SSS similarity with correct correspondence.
Triangle angles add to 180°. Similar figures have the same shape, with a common scale factor for corresponding lengths.
Match angles first
If ∠A = ∠P and ∠B = ∠Q, then ∠C = ∠R because each triangle totals 180°.
Write the order carefully
Write △ABC ∼ △PQR so A matches P, B matches Q and C matches R.
Match side ratios
The corresponding ratios are AB/PQ = BC/QR = AC/PR. Do not mix unmatched sides.
Choose a valid criterion
AA uses two equal angles; SSS uses three proportional side pairs; SAS uses two proportional pairs and the equal included angle.
Worked example: Triangles with side lengths 3, 4, 5 and 6, 8, 10 are similar by SSS because every larger/smaller ratio is 2. They are not congruent because corresponding lengths differ.
Watch for this: Two sides in proportion alone are not enough for SAS; the angle between those sides must match.
Discuss: Draw a triangle and a copy enlarged by factor 3. Label matching vertices and write all three corresponding side ratios.
LESSON 2
Use a parallel line inside a triangle
Apply the basic proportionality theorem and its converse.
A line parallel to one side creates equal corresponding angles.
State the geometry
D lies on AB, E lies on AC and DE is parallel to BC.
Use the divided-side ratio
The theorem gives AD/DB = AE/EC. With AD = 3, DB = 2 and AE = 6, write 3/2 = 6/EC.
Solve the proportion
Cross-multiply: 3EC = 12, so EC = 4.
Check and distinguish the converse
3/2 = 6/4 checks. Conversely, if points on the two sides divide them equally in this ratio, the converse theorem allows us to conclude DE ∥ BC.
Worked example: If AD = 4, DB = 6, AE = 2 and EC = 3, then AD/DB = 2/3 = AE/EC. With D and E on the two sides, the converse theorem gives DE ∥ BC.
Watch for this: AD/AB = AE/AC is also valid in this parallel configuration, but do not mix AD/DB with AE/AC.
Discuss: Draw the diagram and change DB while keeping AD fixed. Predict whether EC must grow or shrink if AE stays fixed.
LESSON 3
Use similarity to measure indirectly
Justify the similar triangles before setting up a scale ratio.
A proportional calculation needs a geometric reason for the proportion.
State assumptions
A vertical 2 m pole and a tree stand on level ground under the same sunlight. The shadows are 3 m and 12 m.
Justify the shape match
Both triangles have right angles, and parallel sunlight gives the same acute sun angle. They are similar by AA.
Match height to shadow
h/12 = 2/3, so h = 12 × 2/3 = 8 m.
Check the result and assumptions
The shadow is four times as long, so the height is four times 2 m. Sloping ground or a leaning pole would require a different model.
Worked example: A 1.5 m reference stick has a 2 m shadow. A wall has a 6 m shadow at the same time. Height = 6 × 1.5/2 = 4.5 m, under the stated level-ground and vertical-object assumptions.
Watch for this: The shadow ratio works only when corresponding angles match. Similarity should be justified, not assumed from a drawing.
Discuss: Use safe tabletop objects and a single lamp to test which conditions are needed before comparing shadows. Record measurements and assumptions.
Original learning companion, not full textbook exercise solutions. Covers correspondence, similarity criteria, BPT applications and indirect measurement. Complete deductive proofs of BPT and all similarity criteria, and every textbook problem, are still outside this companion.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
