IP EMaths Notes (Upper Sec, Year 3-4): 12) Similarity and Congruency

Similarity compares shapes with proportional sides and equal angles, while congruency proves exact matches. Keep diagrams annotated with tick marks, matching angle symbols, and explicit ratio statements so the reader can follow the logic without guessing.

Quick reference

• Similarity shortcut: write \( \triangle ABC \sim \triangle DEF \) when all corresponding angles match and the three side ratios are equal.
• Triangle similarity criteria:
  • \( \text{AAA} \): three angle matches force similarity.
  • \( \text{SAS} \) similarity: a pair of equal angles sandwiched by proportional sides.
  • \( \text{SSS} \) similarity: three side ratios equal.
• Congruency criteria (exact match): \( \text{SSS} \), \( \text{SAS} \), \( \text{ASA} \), \( \text{AAS} \), and right-angle cases \( \text{RHS} \) / \( \text{HL} \).
• Scaling facts for \( \triangle ABC \sim \triangle DEF \):
  • Side ratio \( k = \tfrac{AB}{DE} = \tfrac{BC}{EF} = \tfrac{CA}{FD} \).
  • Perimeter ratio equals \( k \).
  • Area ratio equals \( k^2 \).
  • For similar solids (cones, pyramids) the volume ratio equals \( k^3 \).

Similarity fundamentals

• Identify corresponding parts carefully (write them in order, e.g. \( A \leftrightarrow D \), \( B \leftrightarrow E \)).
• Parallel lines create equal alternate or corresponding angles, a common trigger for similarity.
• Once similarity is confirmed, state both the angle relationships and the chosen side ratio before solving for unknowns.

Congruency fundamentals

• Congruency requires explicit evidence that size and shape match: show equal sides/angles or cite a standard test (e.g. \( \text{SAS} \)).
• Use congruency to transfer results: if \( \triangle ABC \cong \triangle DEF \), then \( AB = DE \), \( \angle C = \angle F \), etc.
• Congruent triangles often lead to symmetry arguments or allow you to justify perpendicular and midpoint statements in geometry proofs.

Worked example - Height via similar triangles

A lamppost casts a shadow \( \pu{6.5 m} \) long when a \( \pu{1.8 m} \) tall student stands nearby and casts a shadow \( \pu{2.0 m} \) long. Assuming the lamppost, student, and shadow tips form similar triangles, find the lamppost height.

1. State similarity: the lamppost triangle and student triangle share the angle of elevation and have right angles at the bases, so \( \triangle\text{Lamp} \sim \triangle\text{Student} \) by \( \text{AA} \).
2. Let \( h \) denote the lamppost height and write the side ratio:
   \[ \frac{h}{\pu{6.5 m}} = \frac{\pu{1.8 m}}{\pu{2.0 m}}. \]
3. Solve for \( h \):
   \[ h = \pu{6.5 m} \times \frac{\pu{1.8 m}}{\pu{2.0 m}} = \pu{6.5 m} \times 0.9 = \pu{5.85 m}. \]
4. Conclude with a sentence: the lamppost is \( \pu{5.85 m} \) tall (to 3 s.f.).

Worked example - Proof from similarity to congruency

Given \( \triangle ABC \) with \( AB = AC \) and point \( D \) on \( AC \) such that \( BD \) is perpendicular to \( AC \). Prove \( \triangle ABD \cong \triangle CBD \) and hence find \( \angle B \).

1. Triangles \( \triangle ABD \) and \( \triangle CBD \) share \( \angle B \) and both contain a right angle at \( D \), so \( \triangle ABD \sim \triangle CBD \) by \( \text{AA} \).
2. Similarity gives side ratio \( \tfrac{AB}{CB} = \tfrac{BD}{BD} = 1 \), so \( AB = CB \).
3. With \( AB = AC \) and \( AB = CB \), triangle \( ABC \) is equilateral, hence \( \angle B = 60^\circ \).
4. Because \( \triangle ABD \) and \( \triangle CBD \) share a side \( BD \) and now have two equal sides, we can tighten to congruency via \( \text{SSS} \). This justifies that \( D \) is the midpoint of \( AC \) and \( BD \) is a perpendicular bisector.

Ratio consequences in practice

• Perimeter scaling: \( \text{Perimeter}_1 = k \times \text{Perimeter}_2 \).
• Area scaling: if \( k = \tfrac{7}{4} \), then \( \text{Area}_1 = k^2 \times \text{Area}_2 = \tfrac{49}{16} \times \text{Area}_2 \).
• When a diagram includes a scale drawing, multiply lengths by the scale factor and areas by its square to convert between model and real sizes.

Try this

1. Two similar triangles have side ratio \( 5:8 \). If the smaller triangle has area \( \pu{45 cm^2} \), find the area of the larger triangle.
2. In \( \triangle PQR \), \( PS \) is drawn parallel to \( QR \) meeting \( PR \) at \( S \). Given \( PS = \pu{4.2 cm} \), \( QR = \pu{9.0 cm} \), and \( PR = \pu{7.0 cm} \), find \( RS \) and the ratio of the areas of \( \triangle PSR \) to \( \triangle PQR \).