Chapter 8.3 — Exercise 8.2 — Similarity Criteria

Three Shortcuts to Similarity

• AA (Angle-Angle): if two angles of one triangle equal two angles of another, the triangles are similar — the third angle is automatically forced to match too, since every triangle's angles sum to 180°. This alone makes AAA collapse into AA: checking three angles is never actually necessary once two already match.
• SSS (Side-Side-Side): if the three sides of one triangle are proportional to the three sides of another, the triangles are similar, and every corresponding angle is automatically equal.
• SAS (Side-Angle-Side): if one angle of a triangle equals one angle of another, and the two sides forming that angle are proportional, the triangles are similar.

All three proofs in this textbook share one underlying construction: mark points P and Q on the second triangle's two relevant sides so that a smaller triangle DPQ is congruent to the first triangle, then show PQ must be parallel to the second triangle's third side — which pulls BPT in to finish the argument. AA, SSS, and SAS aren't three unrelated facts; they're three different starting conditions that all funnel through the identical proof machinery. Whichever criterion a problem hands over — two matching angles, three proportional sides, or one matching angle with two proportional sides around it — the underlying justification for calling the triangles similar is always this same construction working in the background, even when a solution only cites "by AA" or "by SAS" without spelling it out again each time.

One more fact falls out for free once similarity is confirmed: the ratio of the two triangles' perimeters equals the same ratio as any pair of corresponding sides — AB/DE = BC/EF = AC/DF = (AB+BC+AC)/(DE+EF+DF).

Spotting Similarity at a Glance

The second-to-last row is the one worth pausing on: matching angles alone (AAA, or AA once the third is inferred) is always enough by itself, no side-length check required at all — a genuinely different bar to clear than SSS or SAS, both of which need actual length ratios confirmed before angles can be assumed equal. The very first row is worth its own comparison too: two ratios that are merely close (2 versus 2.5) still fail the test completely — similarity needs the ratios exactly equal, not approximately equal, however small the gap looks.

Finding an Unknown Side From a Given Similarity

Once two triangles are already known to be similar, an unknown side falls out of a single proportion:

Combining a Proof With a Calculation

Given ∠ADE = ∠B in △ABC (D on AB, E on AC): first show △ABC ~ △ADE, then use that similarity to find DE if AD=3.8, AE=3.6, BE=2.1, BC=4.2.

With AB = AE+BE = 5.7, substituting gives 5.7/3.8 = 4.2/DE, and since 3.8/5.7 simplifies exactly to 2/3, DE = 4.2×(2/3) = 2.8 cm. The proof step isn't just a formality here — without confirming AA similarity first, there'd be no justified proportion to substitute numbers into at all. It's worth noticing, too, that AB itself was never given directly — it had to be built first from AE+BE, a small assembly step easy to skip past without noticing it's even necessary.

A Perimeter Ratio, No Angles Needed

Two similar triangles have perimeters 30 cm and 20 cm; one side of the first triangle is 12 cm. Since the perimeter ratio always equals the corresponding-side ratio, the matching side in the second triangle is found without ever touching an angle: 12/x = 30/20, giving x = 8 cm.

Two Real-World Shadows

Both of this exercise's word problems use the same underlying triangle pair — an object, its shadow, and the parallel sun rays that make the object-and-shadow triangle similar to a second, larger or smaller one nearby.

• A girl's shadow: a 90 cm girl walks away from a 3.6 m lamp post at 1.2 m/s. After 4 seconds she's walked 4.8 m. With similar triangles formed by the lamp post/girl and their shared shadow line, 3.6/0.9 = (x+4.8)/x simplifies to 4x=x+4.8, giving x = 1.6 m of shadow.
• A flagpole and a building: a 4 m flagpole casts a 6 m shadow at the same moment a nearby building casts a 24 m shadow. Since both shadows are cast under the same sun angle, 4/6 = x/24, giving x = 16 m for the building's height.

Two Structural Proofs Worth Comparing

A trapezium ABCD with AB ∥ DC has diagonals meeting at O. Unlike the version of this same problem in Exercise 8.1, which needed an auxiliary line through O, this exercise proves the identical result — AO/OC = BO/OD — directly: △OAB ~ △OCD by AA (vertically opposite angles at O, alternate interior angles from AB∥DC), and similarity alone hands over the ratio immediately. Two entirely different toolkits reaching the same destination is worth noticing on its own.

A second proof chains one given similarity into two new ones: if △ABC~△PQR with medians CM and RN, then since M and N are midpoints, AM/PN = AB/PQ = AC/PR automatically — enough for SAS similarity to give △AMC~△PNR, and from there CM/RN = AB/PQ falls out directly. The same reasoning, mirrored on the other half of each triangle, gives a third similarity, △CMB~△RNQ. One given fact about the whole triangles produces two new facts about their medians, entirely through substitution.

Not Every Problem Here Is a Proof or a Number

Three problems in this exercise ask for something else entirely: drawing a triangle similar to a given one at a stated scale factor (5/3 times, 2/3 times, 1½ times), using ratio-based compass construction rather than any calculation. Like Exercise 8.1's ratio-division construction, these test a different, hands-on skill — accurately transferring a scale factor onto paper — that no amount of algebra substitutes for.

From Matching Shapes to Matching Areas

Every similarity fact in this exercise compared lengths — sides, perimeters, medians. Exercise 8.3 asks what happens to area once two triangles are already known to be similar, and the answer turns out to involve squaring the same ratio this exercise spent its time establishing. For the AA/SSS/SAS criteria this exercise's every proof depends on, see the sections above, or revisit BPT itself in the chapter introduction.