Triangles

Two photographs of the Taj Mahal — one postcard-sized, one poster-sized — are not the same size, yet everyone agrees they show the same monument in the same shape. That everyday idea, sameness of shape without sameness of size, is what mathematics calls similarity, and this chapter turns it into a precise, provable toolkit. You already know congruence from Class IX: figures that match exactly in both shape and size. Similarity is the more general, more useful cousin — it asks only that corresponding angles be equal and corresponding sides be in the same ratio, and it is similarity, not congruence, that lets a surveyor find the height of a mountain or the width of a river without ever touching it. The chapter builds in a deliberate order. It starts with similar polygons in general, then narrows to triangles, where two remarkable theorems appear: the Basic Proportionality Theorem (Thales' theorem), which says a line parallel to one side of a triangle slices the other two sides in the same ratio, and its converse. These feed directly into the four criteria that let you prove two triangles are similar — AAA/AA, SSS, and SAS similarity — each one economising on how much information you actually need to check. Once similarity is established between two triangles, a cascade of results follows almost for free: corresponding medians, altitudes, and angle bisectors all inherit the same ratio as the corresponding sides, and — though NCERT keeps the deepest results (areas of similar triangles, the Pythagoras theorem as a consequence of similarity) for the chapter's exercises and the next chapter — the seeds of that connection are planted here. By the end of this chapter you should be able to look at two triangles, decide in seconds which similarity test applies, set up the correct ratio equation, and solve for an unknown side, angle, or distance — the exact skill tested in board exams through indirect-measurement word problems (shadows, poles, ponds) and proof-based questions built entirely on the AA/SSS/SAS criteria.

Key takeaways

Similar figures have the same shape but not necessarily the same size; congruent figures are always similar, but similar figures need not be congruent.
Two polygons of the same number of sides are similar only if BOTH conditions hold: corresponding angles equal AND corresponding sides proportional. Either condition alone is not enough.
Basic Proportionality Theorem (Thales, Theorem 6.1): if a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, it divides those two sides in the same ratio: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ .
Converse of BPT (Theorem 6.2): if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
AAA/AA similarity (Theorem 6.3): if corresponding angles of two triangles are equal, the triangles are similar (and just two equal angles are enough, since the third follows from the angle-sum property).
SSS similarity (Theorem 6.4): if corresponding sides of two triangles are all in the same ratio, the triangles are similar.
SAS similarity (Theorem 6.5): if one angle of a triangle equals one angle of another and the sides including those angles are proportional, the triangles are similar.
For similar triangles, the ratio of any pair of corresponding sides equals the ratio of corresponding medians, corresponding altitudes, and corresponding angle bisectors.
Similarity symbol is $\sim$ ; correspondence of vertices matters — $\triangle ABC \sim \triangle DEF$ is a different statement from $\triangle ABC \sim \triangle EDF$ unless the angles actually match that way.
Key idea: two triangles are similar exactly when their corresponding angles match (AA is enough) or their corresponding sides are all in one common ratio (SSS) or one angle matches with its including sides in ratio (SAS), and the Basic Proportionality Theorem is the engine that proves all of it.

The core insight

Similar triangles are triangles with the same shape (equal corresponding angles, proportional corresponding sides), and the Basic Proportionality Theorem — a line parallel to one side cuts the other two sides in the same ratio — is the single mechanism behind the AA, SSS and SAS similarity criteria used to prove it.

What this chapter covers

The chapter has five moving parts, and each is a prerequisite for the next. Section 6.2 defines similar figures in general (any polygons) and states the twin conditions — equal angles, proportional sides — that similarity demands. Section 6.3 narrows to similarity of triangles and proves the Basic Proportionality Theorem and its converse, the workhorse results used to set up almost every ratio equation in this chapter. Section 6.4 delivers the criteria for similarity of triangles — AAA/AA, SSS, SAS — each with a proof built on cutting off a smaller triangle and using BPT. Worked examples throughout show these criteria applied to real configurations: parallel lines, vertically opposite angles, shadows cast by lamp-posts, medians of similar triangles. A closing note (outside the formal theorem list but worth knowing) mentions the RHS similarity criterion for right triangles. Master the four boxed theorems and you can answer essentially every proof-based question this chapter sets.

The core ideas

Similar figures and the two-condition test

Two figures are congruent if they have the same shape and the same size; they are similar if they have the same shape but not necessarily the same size. Every congruent figure is automatically similar (scale factor $1$ ), but similar figures need not be congruent — a passport-size and a postcard-size photo of the same scene are similar, not congruent. Two polygons with the same number of sides are similar precisely when both of the following hold together:

\angle A = \angle A', \ \angle B = \angle B', \ \ldots \qquad \text{and} \qquad \frac{AB}{A'B'} = \frac{BC}{B'C'} = \cdots \qquad (1)

where: the primed letters denote the corresponding vertices of the second polygon, and the common value of the side ratios is called the scale factor (or representative fraction). Neither condition alone suffices: a square and a rectangle have all angles equal ( $90°$ each) but sides not in one common ratio, so they are not similar; a square and a rhombus can have sides in the same ratio but unequal angles, so they too fail to be similar. Both conditions must hold simultaneously.

Similarity of triangles and the correspondence notation

A triangle is a polygon, so the same two-condition definition applies, but for triangles the two conditions turn out to be equivalent to each other (proved in the next section), which is what makes triangles special and gives rise to shortcut criteria. When $\triangle ABC$ is similar to $\triangle DEF$ , written $\triangle ABC \sim \triangle DEF$ , the correspondence of vertices is part of the statement: $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ , so

\angle A = \angle D, \ \angle B = \angle E, \ \angle C = \angle F, \qquad \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}. \qquad (2)

where: the order of letters fixes which vertex maps to which; writing $\triangle ABC \sim \triangle EDF$ for the same pair of triangles would be a different (generally false) claim unless the angles genuinely matched that way. Getting this order right is what lets you read off the correct ratio equation instantly from a similarity statement — a skill examiners test directly.

Figure

A triangle ABC with a line DE drawn parallel to side BC cutting AB at D and AC at E illustrating the Basic Proportionality Theorem

Reading a similarity criterion off a figure

Given two triangles, you rarely have all six pairs of corresponding parts (three angles, three sides) measured. The three criteria below are exactly the minimum information needed. Written compactly, they are:

\text{AA: } \angle A = \angle D,\ \angle B = \angle E \ \Rightarrow\ \triangle ABC \sim \triangle DEF; \quad \text{SSS: } \frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD} \ \Rightarrow\ \triangle ABC \sim \triangle DEF; \quad \text{SAS: } \frac{AB}{DE}=\frac{AC}{DF},\ \angle A = \angle D \ \Rightarrow\ \triangle ABC \sim \triangle DEF. \qquad (3)

where: AA needs only two angle equalities (the third follows from the angle-sum property, $180°$ total), SSS needs all three side ratios equal but no angle information at all, and SAS needs one angle plus the two sides that include that angle — an angle that is not between the two given sides does not qualify. Choosing the right criterion for a given figure, and checking that the angle in SAS really is the included angle, is the single most common source of marks lost in board proofs.

Figure

Three pairs of triangles side by side illustrating the AA SSS and SAS similarity criteria with matching tick marks and angle arcs

Derivations

Derivation AThe Basic Proportionality Theorem and its proof by areas

The goal is to prove Theorem 6.1: if $DE$ is drawn parallel to side $BC$ of $\triangle ABC$ , meeting $AB$ at $D$ and $AC$ at $E$ , then $D$ and $E$ divide the two sides in the same ratio, $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ . The proof is elegant because it needs nothing beyond the area formula "half base times height" and one observation about triangles sharing a base between the same parallels — no trigonometry, no coordinates, just areas compared four different ways.

Step 1. Join $BE$ and $CD$ , and drop perpendiculars $EN$ to $AB$ and $DM$ to $AC$ . This gives four small triangles inside the figure — $\triangle ADE$ , $\triangle BDE$ , $\triangle DEC$ — each of whose areas can be written using a side of $\triangle ADE$ as base.

Step 2. Compute two pairs of areas using $AD$ and $DB$ as bases (both triangles share the same height $EN$ from $E$ to line $AB$ ): $\text{ar}(\triangle ADE) = \tfrac12 \cdot AD \cdot EN$ and $\text{ar}(\triangle BDE) = \tfrac12 \cdot DB \cdot EN$ . Dividing, the common factor $\tfrac12 EN$ cancels, giving $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{AD}{DB}$ .

Step 3. Similarly, using $AE$ and $EC$ as bases with the shared height $DM$ from $D$ to line $AC$ : $\text{ar}(\triangle ADE) = \tfrac12 \cdot AE \cdot DM$ and $\text{ar}(\triangle DEC) = \tfrac12 \cdot EC \cdot DM$ , so $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \dfrac{AE}{EC}$ .

Step 4. The key extra fact: $\triangle BDE$ and $\triangle DEC$ sit on the same base $DE$ and lie between the same two parallel lines $DE$ and $BC$ (since $DE \parallel BC$ by hypothesis). Triangles on the same base between the same parallels have equal area, so $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$ .

Step 5. Combine the three results. Since the two ratios from Steps 2 and 3 have equal denominators (Step 4), the numerators' ratio must match too: $\dfrac{AD}{DB} = \dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \dfrac{AE}{EC}$ . This proves $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , exactly as claimed, and by adding $1$ to both sides one immediately also gets the equivalent form $\dfrac{AB}{AD} = \dfrac{AC}{AE}$ , used constantly in worked problems.

Go deeper · Derivation B · 1 minThe AAA similarity criterion, proved using the Basic Proportionality Theorem

Derivation BThe AAA similarity criterion, proved using the Basic Proportionality Theorem

The goal is to prove Theorem 6.3: if $\angle A = \angle D$ , $\angle B = \angle E$ , $\angle C = \angle F$ for triangles $ABC$ and $DEF$ , then the corresponding sides are automatically in the same ratio, so $\triangle ABC \sim \triangle DEF$ . The strategy is a classic "cut and compare": carve out of the larger triangle a smaller one congruent to $\triangle ABC$ , show that smaller triangle's third side is parallel to the base, and then let the Basic Proportionality Theorem do the rest of the work.

Step 1. Without loss of generality suppose $\triangle DEF$ is at least as large as $\triangle ABC$ (i.e. $DE \geq AB$ ). On ray $DE$ mark point $P$ with $DP = AB$ , and on ray $DF$ mark point $Q$ with $DQ = AC$ ; join $PQ$ . By construction $DP = AB$ , $DQ = AC$ , and $\angle PDQ = \angle A = \angle D$ (the given angle equality), so by the SAS congruence criterion from Class IX, $\triangle ABC \cong \triangle DPQ$ .

Step 2. Congruent triangles have equal corresponding angles, so $\angle DPQ = \angle ABC = \angle B$ . But we are given $\angle B = \angle E$ , hence $\angle DPQ = \angle E$ . Since $\angle DPQ$ and $\angle DEF$ (that is, $\angle E$ ) are corresponding angles for the two lines $PQ$ and $EF$ cut by the transversal $DE$ , their equality forces $PQ \parallel EF$ .

Step 3. Now apply the Basic Proportionality Theorem (proved above) to $\triangle DEF$ with the line $PQ$ drawn parallel to side $EF$ : it gives $\dfrac{DP}{PE} = \dfrac{DQ}{QF}$ . Adding $1$ to each side, $\dfrac{DE}{DP} = \dfrac{DF}{DQ}$ , i.e. $\dfrac{DP}{DE} = \dfrac{DQ}{DF}$ .

Step 4. Substitute back $DP = AB$ and $DQ = AC$ from Step 1: $\dfrac{AB}{DE} = \dfrac{AC}{DF}$ . Repeating the identical construction with a different pair of vertices (cutting off a triangle equal to $\triangle ABC$ from the $\angle E$ corner instead) gives, by the same argument, $\dfrac{AB}{DE} = \dfrac{BC}{EF}$ . Combining, $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$ — corresponding sides are in one common ratio, so together with the given equal angles, $\triangle ABC \sim \triangle DEF$ , exactly the definition of similarity. Since equal angles in two triangles automatically force the third pair equal (angle-sum property, $180°$ ), checking just two angle pairs — the AA shortcut — is enough to invoke this whole theorem.

Go deeper · Derivation C · 1 minThe SAS similarity criterion, proved by the same cut-and-parallel method

Derivation CThe SAS similarity criterion, proved by the same cut-and-parallel method

The goal is to prove Theorem 6.5: if $\angle A = \angle D$ and the two sides including that angle are proportional, $\dfrac{AB}{DE} = \dfrac{AC}{DF}$ , then $\triangle ABC \sim \triangle DEF$ . This matters because it lets you certify similarity from just one angle and two sides — no need to measure or compute the third side or the other two angles at all — and it is the criterion used most often when a problem hands you vertically opposite or alternate angles together with a couple of given lengths.

Step 1. As before, on ray $DE$ cut $DP = AB$ and on ray $DF$ cut $DQ = AC$ , then join $PQ$ . Since $DP = AB$ , $DQ = AC$ , and $\angle PDQ = \angle A = \angle D$ by hypothesis, SAS congruence (Class IX) gives $\triangle ABC \cong \triangle DPQ$ immediately — a full triangle match, not just a similarity, at this intermediate stage.

Step 2. From the hypothesis $\dfrac{AB}{DE} = \dfrac{AC}{DF}$ and the substitutions $AB = DP$ , $AC = DQ$ , we get $\dfrac{DP}{DE} = \dfrac{DQ}{DF}$ . This is precisely the "line divides two sides in the same ratio" condition of the converse Basic Proportionality Theorem (Theorem 6.2), applied inside $\triangle DEF$ to the point $P$ on $DE$ and $Q$ on $DF$ .

Step 3. By the converse BPT, therefore, $PQ \parallel EF$ . Two lines cut by transversal $DE$ with $PQ \parallel EF$ force equal corresponding angles: $\angle DPQ = \angle DEF$ and $\angle DQP = \angle DFE$ .

Step 4. From Step 1, congruence already gave $\angle DPQ = \angle ABC = \angle B$ and $\angle DQP = \angle ACB = \angle C$ (corresponding angles of congruent triangles $\triangle ABC$ and $\triangle DPQ$ ). Combining with Step 3, $\angle B = \angle DPQ = \angle E$ and $\angle C = \angle DQP = \angle F$ . Together with the given $\angle A = \angle D$ , all three pairs of angles of $\triangle ABC$ and $\triangle DEF$ are now equal, and the given side ratio extends to the third pair by the AAA argument above. Hence $\triangle ABC \sim \triangle DEF$ , completing the SAS proof.

Worked examples

Board level - proving a ratio from a parallel line (NCERT Example 1 style)

A line intersects sides $AB$ and $AC$ of $\triangle ABC$ at $D$ and $E$ respectively and is parallel to $BC$ . Prove that $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ .

Step 1. Since $DE \parallel BC$ , the Basic Proportionality Theorem applies directly: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ .

Step 2. Take reciprocals of both sides (valid since none of the lengths is zero): $\dfrac{DB}{AD} = \dfrac{EC}{AE}$ .

Step 3. Add $1$ to each side: $\dfrac{DB}{AD} + 1 = \dfrac{EC}{AE} + 1$ , i.e. $\dfrac{DB + AD}{AD} = \dfrac{EC + AE}{AE}$ .

Step 4. Since $D$ lies on $AB$ , $DB + AD = AB$ ; since $E$ lies on $AC$ , $EC + AE = AC$ . Substituting, $\dfrac{AB}{AD} = \dfrac{AC}{AE}$ , and taking reciprocals once more gives the required result. Answer: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ , proved.

Board level - the lamp-post shadow problem (NCERT Example 7 style)

A girl of height $90\ \text{cm}$ is walking away from the base of a lamp-post at a speed of $1.2\ \text{m/s}$ . The lamp is $3.6\ \text{m}$ above the ground. Find the length of her shadow after $4$ seconds.

Step 1. Convert units and set up the figure: $AB$ is the lamp-post ( $AB = 3.6\ \text{m}$ ), $CD$ is the girl ( $CD = 90\ \text{cm} = 0.9\ \text{m}$ ), and after walking for $4$ seconds she has moved $BD = 1.2 \times 4 = 4.8\ \text{m}$ from the lamp-post's foot. Let $DE = x$ metres be the length of her shadow, so $E$ is the tip of the shadow on the ground and $BE = BD + DE = 4.8 + x$ .

Step 2. Identify the similar triangles: in $\triangle ABE$ and $\triangle CDE$ , $\angle B = \angle D = 90°$ (both the lamp-post and the girl stand vertical to the ground) and $\angle E = \angle E$ (the same angle, shared vertex). By the AA similarity criterion, $\triangle ABE \sim \triangle CDE$ .

Step 3. Corresponding sides are proportional: $\dfrac{AB}{CD} = \dfrac{BE}{DE}$ , i.e. $\dfrac{3.6}{0.9} = \dfrac{4.8+x}{x}$ .

Step 4. Simplify the left side: $\dfrac{3.6}{0.9} = 4$ , so $4 = \dfrac{4.8+x}{x}$ , giving $4x = 4.8 + x$ , so $3x = 4.8$ , so $x = 1.6$ . Answer: the shadow is $\mathbf{1.6\ \text{m}}$ long.

Board HOTS / foundation JEE-NEET - similar triangles from similar triangles' medians (NCERT Example 8 style)

$CM$ and $RN$ are respectively the medians of $\triangle ABC$ and $\triangle PQR$ . If $\triangle ABC \sim \triangle PQR$ , prove that (i) $\triangle AMC \sim \triangle PNR$ , and (ii) $\dfrac{CM}{RN} = \dfrac{AB}{PQ}$ .

Step 1. From the given $\triangle ABC \sim \triangle PQR$ , extract both pieces of the similarity: the angles $\angle A = \angle P$ , $\angle B = \angle Q$ , $\angle C = \angle R$ , and the side ratios $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$ .

Step 2. Because $CM$ and $RN$ are medians, $M$ and $N$ are midpoints, so $AB = 2\,AM$ and $PQ = 2\,PN$ . Substitute into the ratio $\dfrac{AB}{PQ} = \dfrac{CA}{RP}$ from Step 1: $\dfrac{2\,AM}{2\,PN} = \dfrac{CA}{RP}$ , which simplifies to $\dfrac{AM}{PN} = \dfrac{CA}{RP}$ .

Step 3. In $\triangle AMC$ and $\triangle PNR$ we now have two sides in ratio ( $\dfrac{AM}{PN} = \dfrac{CA}{RP}$ , from Step 2) and the angle between them equal ( $\angle MAC = \angle A = \angle P = \angle NPR$ , from Step 1, since $M$ lies on $AB$ and $N$ lies on $PQ$ so $\angle MAC$ is the same angle as $\angle BAC$ ). This is exactly the SAS similarity criterion, so $\triangle AMC \sim \triangle PNR$ , proving part (i).

Step 4. From part (i), corresponding sides of the newly similar triangles are proportional: $\dfrac{CM}{RN} = \dfrac{CA}{RP}$ . But Step 1 already gave $\dfrac{CA}{RP} = \dfrac{AB}{PQ}$ (both equal the common ratio of $\triangle ABC \sim \triangle PQR$ ). Chaining the two equalities, $\dfrac{CM}{RN} = \dfrac{AB}{PQ}$ . Answer: both parts proved — corresponding medians of similar triangles are in the same ratio as the corresponding sides. Timing note: aim for under 5 minutes on a part-(i)+(ii) proof like this. The reusable pattern is: pull both the angle-equalities and the side-ratio out of the given similarity FIRST (Step 1), then hunt for which SAS/SSS/AA setup the sub-triangles satisfy — medians and altitudes almost always resolve via SAS because "half of a proportional side" is still proportional. Do not try to prove (ii) from scratch; it falls out of (i) in one line, so proving (i) correctly is where all the marks live.

Common traps

Students most often fail on correspondence order: writing $\triangle ABC \sim \triangle EFD$ when the angles actually match as $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ produces a false statement even if "the triangles are similar" is true — order encodes which sides correspond, and getting it wrong flips every subsequent ratio. A second trap is applying SAS similarity with the wrong angle: the equal angle must be the one included between the two sides used in the ratio, not any angle of the two triangles. A third is forgetting that AAA collapses to AA — two angle equalities are sufficient and citing all three is not wrong but wastes time verifying something the angle-sum property already guarantees. A fourth is misreading the Basic Proportionality Theorem's ratio direction — it is $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ (each divided segment over its remaining partner on the same side), not $\dfrac{AD}{AB}$ , though the equivalent "whole side" form $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ is also valid and often more convenient — mixing the two forms mid-calculation is a common source of arithmetic slips. Finally, remember that either condition of similarity (equal angles, or proportional sides) taken alone is not sufficient for general polygons — it is a triangle-specific theorem (proved above) that the two conditions become equivalent; do not assume it for quadrilaterals or other polygons.

Common trapSimilar does NOT mean congruent, and equal angles alone do not prove similarity for non-triangle polygons

The single most damaging mix-up in this chapter is treating "similar" as a weaker synonym for "congruent," or assuming that matching angles is by itself enough to prove two figures similar. For triangles specifically, equal corresponding angles DO force proportional sides (that is the content of the AAA theorem, proved above) — but for any polygon with four or more sides, equal angles alone (as in a square versus a rectangle) do not guarantee similarity, and proportional sides alone (as in a square versus a rhombus) do not either; both conditions must be checked together. Keep the vocabulary precise: congruent figures are always similar (scale factor exactly $1$ ), but similar figures are congruent only in the special case their scale factor happens to be $1$ .

You should be able to…

State precisely when two polygons (and specifically two triangles) are similar, using both the angle condition and the side-ratio condition together.
Explain why congruent figures are always similar but similar figures need not be congruent, with an example of each.
State the Basic Proportionality Theorem and its converse, and know which one to invoke based on whether you are given a parallel line or given a ratio.
Prove the Basic Proportionality Theorem from scratch using the area method (same base, same parallels).
Identify which of AA, SSS or SAS similarity applies to a given pair of triangles from the information supplied, and set up the correct correspondence-respecting ratio equation.
Solve indirect-measurement word problems (shadows, poles, heights) by identifying the two similar right triangles and equating ratios.
Prove that corresponding medians (or altitudes, or angle bisectors) of similar triangles are in the same ratio as the corresponding sides.
Correctly write similarity statements with vertices in the matching order, and read off the correct side ratios from that order.
Recognise the two classic error patterns: assuming any one condition alone (angles or sides) proves similarity for general polygons, and using SAS with the wrong (non-included) angle.

Ask the tutor: Want to work through a board-style similarity proof from your textbook together, step by step?

What exactly is the difference between congruent and similar triangles?

Congruent triangles are identical in every respect — same three angles and same three side lengths, so one can be picked up and placed exactly on top of the other. Similar triangles only need the same shape: equal corresponding angles and sides that are all in one common ratio (the scale factor), but the sizes can differ. Every congruent pair is automatically similar, with scale factor exactly $1$ ; but two similar triangles are congruent only in that special case. Board exams usually cue you: if the question mentions ratios or asks you to "prove similar," reach for AA/SSS/SAS; if it gives you equal side lengths outright and asks to "prove congruent," reach for SSS/SAS/ASA/RHS congruence from Class IX instead.

Why is AA enough for similarity — don't we need all three angles to match?

Because of the angle-sum property of a triangle: the three interior angles of any triangle always add to $180°$ . If you already know two angles of $\triangle ABC$ equal two angles of $\triangle DEF$ , the third angle of each triangle is forced to be $180°$ minus the sum of the other two — and since those two sums are equal, the third angles must be equal too. So checking a third pair adds no new information; it is guaranteed once the first two match. This is exactly why the theorem is often just called "AA similarity" even though the formal name is AAA.

How do I know which similarity criterion to use in a given problem?

Look at what the problem actually hands you. If two angles are stated or derivable (often from parallel lines giving alternate/corresponding angles, or vertically opposite angles at an intersection), reach for AA. If you are given all three side lengths (or ratios) for both triangles and no angle information, use SSS. If you are given exactly one angle plus the two sides that include it, use SAS — but double check the angle is genuinely between those two given sides, since an angle opposite one of them will not work. Most board problems are engineered to make exactly one criterion fit cleanly; if your first attempt needs information you were not given, you have probably picked the wrong one.

Does the Basic Proportionality Theorem work if the line is not parallel to the side?

No — parallelism is the entire hypothesis of Theorem 6.1, and without it the two sides need not be divided in the same ratio at all. This is exactly why the converse (Theorem 6.2) is a genuinely separate, useful theorem in its own right: it lets you go the other way, starting from a proportional division and concluding the line must be parallel, which is how many "prove EF is parallel to QR" problems in Exercise 6.2 are actually solved. Whenever a problem states or implies a line is parallel to a triangle's side, BPT is very likely the fastest route to the ratio you need; whenever it gives you a ratio and asks you to prove parallelism, reach for the converse instead.