Triangles
Why This Matters
How did anyone ever measure the height of Mount Everest, or the distance to the Moon, without climbing it or flying there with a tape measure? You can’t. Yet we know both numbers. The trick is indirect measurement — and it rests on one beautiful idea: figures that have the same shape behave in predictable, proportional ways, even when their sizes are wildly different.
A photograph of the Taj Mahal in stamp size and the same photo in postcard size are clearly the same shape — every length has just been scaled up by the same factor, and every angle is unchanged. That “same shape, different size” relationship is called similarity, and triangles are where it becomes a precision tool.
Once you can say “this small triangle is similar to that huge one,” you can turn a measurable length (the small triangle) into an unmeasurable one (the big one) just by keeping the ratios equal. A 6 m pole and its 4 m shadow tell you the height of a tower from its shadow. The same logic powers map scales, blueprints, and the proof of the Pythagoras theorem. This chapter builds that tool from the ground up.
The Big Idea
Two figures are similar if they have the same shape — for polygons that means corresponding angles are equal and corresponding sides are in the same ratio. For triangles, this idea becomes powerful and economical: you don’t have to check all six things. The Basic Proportionality Theorem says a line parallel to one side of a triangle cuts the other two sides in the same ratio — and from it flow simple shortcuts (AA, SSS, SAS) that let you declare two triangles similar after checking just three facts. Similar triangles then let you compute lengths you could never measure directly.
Let’s Break It Down
Similar figures and similar triangles
Recall from Class IX: two figures are congruent if they have the same shape and the same size. Similarity keeps the “same shape” but drops the “same size”.
- All circles are similar. All squares are similar. All equilateral triangles are similar.
- Every congruent pair is similar, but a similar pair need not be congruent.
For polygons with the same number of sides, “same shape” is captured by two conditions together:
- Corresponding angles are equal, and
- Corresponding sides are in the same ratio (proportion).
That common ratio is called the scale factor. Crucially, both conditions are needed for polygons. A square and a rhombus have proportional sides (all equal) but unequal angles — not similar. A square and a rectangle have equal angles but non-proportional sides — also not similar.
For triangles we write similarity with the symbol ∼. When we say △ABC ∼ △DEF, the order of letters matters — it tells you the correspondence: A↔D, B↔E, C↔F. So it means
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD.
Are all isosceles triangles similar to one another?
The Basic Proportionality Theorem (Thales’ Theorem)
This is the engine of the whole chapter.
Theorem 6.1 (BPT). If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: A triangle ABC with a line parallel to BC meeting AB at D and AC at E. To prove: AD/DB = AE/EC.
Construction: Join BE and CD. From E drop EN ⊥ AB, and from D drop DM ⊥ AC.
Proof. Area of a triangle is ½ × base × height.
Taking AD as base and EN as the corresponding height: ar(ADE) = ½ × AD × EN. Taking DB as base and the same height EN: ar(BDE) = ½ × DB × EN.
So ar(ADE) / ar(BDE) = (½ × AD × EN) / (½ × DB × EN) = AD/DB … (1)
Now take the other pair of heights. Taking AE as base and DM as height: ar(ADE) = ½ × AE × DM, and taking EC as base with the same height DM: ar(DEC) = ½ × EC × DM. So
ar(ADE) / ar(DEC) = AE/EC … (2)
Here is the key observation: △BDE and △DEC stand on the same base DE and lie between the same parallels DE and BC. Triangles on the same base and between the same parallels have equal areas, so
ar(BDE) = ar(DEC) … (3)
From (1), (2) and (3), the two left-hand sides ar(ADE)/ar(BDE) and ar(ADE)/ar(DEC) are equal (same numerator, equal denominators), hence
AD/DB = AE/EC. ∎
In △ABC, D lies on AB and E on AC with DE ∥ BC. If AD = 1.5 cm, DB = 3 cm and AE = 1 cm, find EC.
- DE ∥ BC, so by the Basic Proportionality Theorem AD/DB = AE/EC.
- Substitute the known lengths: 1.5/3 = 1/EC.
- So 1.5 × EC = 3 × 1 → EC = 3/1.5.
- Therefore EC = 2 cm.
The converse of the BPT
The reverse is also true — and just as useful, because it lets you prove a line is parallel just from a ratio.
Theorem 6.2 (Converse of BPT). If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Why it holds. Suppose in △ABC a line meets AB at D and AC at E with AD/DB = AE/EC, but DE is not parallel to BC. Then draw a different line DE′ through D that is parallel to BC, with E′ on AC. By the BPT (Theorem 6.1), AD/DB = AE′/E′C. Comparing with the given AD/DB = AE/EC, we get AE/EC = AE′/E′C. Add 1 to both sides:
AE/EC + 1 = AE′/E′C + 1 → (AE + EC)/EC = (AE′ + E′C)/E′C → AC/EC = AC/E′C.
So EC = E′C, which forces E and E′ to coincide. But that means DE is the same line as DE′, which was parallel to BC. So DE ∥ BC. ∎
In △PQR, S lies on PQ and T on PR with PS/SQ = PT/TR, and ∠PST = ∠PRQ. Prove that △PQR is isosceles.
- Since PS/SQ = PT/TR, the line ST divides PQ and PR in the same ratio. By the converse of the BPT, ST ∥ QR.
- ST ∥ QR gives ∠PST = ∠PQR (corresponding angles). … (i)
- We are also given ∠PST = ∠PRQ. … (ii). From (i) and (ii), ∠PQR = ∠PRQ.
- Sides opposite equal angles are equal, so PR = PQ. Hence △PQR is isosceles.
Criteria for similarity of triangles
Checking all six conditions (three angles + three side ratios) is exhausting. For triangles, just three facts are enough. There are three criteria, and each is a theorem the chapter establishes.
Theorem 6.3 — AAA (and hence AA). If in two triangles the corresponding angles are equal, then their corresponding sides are in the same ratio, and the triangles are similar.
Proof sketch. Let △ABC and △DEF have ∠A = ∠D, ∠B = ∠E, ∠C = ∠F. On DE and DF, cut DP = AB and DQ = AC, and join PQ. Then △ABC ≅ △DPQ (SAS: DP = AB, DQ = AC, included ∠D = ∠A). So ∠DPQ = ∠B = ∠E, which makes PQ ∥ EF (equal corresponding angles). By the BPT applied in △DEF, DP/PE = DQ/QF, and adding 1 to each gives DE/DP = DF/DQ, i.e. AB/DE = AC/DF. Repeating with another pair of sides gives AB/DE = BC/EF = AC/DF. Hence the triangles are similar. ∎
Because the angle sum of a triangle is 180°, if two angles match, the third must too. So AAA is usually used as AA: two pairs of equal angles is enough.
Theorem 6.4 — SSS. If the sides of one triangle are proportional to the sides of another, then their corresponding angles are equal and the triangles are similar.
Proof sketch. Suppose AB/DE = BC/EF = CA/FD. Cut DP = AB and DQ = AC on DE, DF and join PQ. From the given ratios one shows DP/PE = DQ/QF, so PQ ∥ EF and DP/DE = DQ/DF = PQ/EF. Combined with the given ratios, PQ = BC. Now △ABC ≅ △DPQ (SSS), so ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, and the triangles are similar. ∎
Theorem 6.5 — SAS. If one angle of a triangle equals one angle of another and the sides including these angles are proportional, then the triangles are similar.
Proof sketch. Let ∠A = ∠D and AB/DE = AC/DF. Cut DP = AB, DQ = AC on DE, DF and join PQ. The side ratios give PQ ∥ EF, so △ABC ≅ △DPQ (SAS) and ∠B = ∠E, ∠C = ∠F. Hence △ABC ∼ △DEF. ∎
| Criterion | What you must show | In short |
|---|---|---|
| AA (from AAA) | Two pairs of corresponding angles equal | two equal angles ⇒ similar |
| SSS | All three pairs of corresponding sides in the same ratio | AB/DE = BC/EF = CA/FD |
| SAS | One pair of equal angles AND the two sides including it proportional | ∠A = ∠D and AB/DE = AC/DF |
Two segments PQ and RS cross at O with PQ ∥ RS. Prove that △POQ ∼ △SOR.
- PQ ∥ RS with PS as a transversal, so ∠P = ∠S (alternate angles). Similarly, with QR as transversal, ∠Q = ∠R.
- At O, ∠POQ = ∠SOR (vertically opposite angles).
- So all three angle pairs are equal — or even just two of them, which is enough.
- By the AA (AAA) similarity criterion, △POQ ∼ △SOR.
In △ABC, AB = 3.8, BC = 6, CA = 3√3, and ∠A = 80°, ∠B = 60°. In △PQR, QP = 12, RQ = 7.6, PR = 6√3. Find ∠P.
- Compare side ratios in the correspondence A↔R, B↔Q, C↔P: AB/RQ = 3.8/7.6 = 1/2, BC/QP = 6/12 = 1/2, CA/PR = 3√3/6√3 = 1/2.
- All three ratios equal 1/2, so by SSS similarity △ABC ∼ △RQP.
- Corresponding angles are equal, so ∠C = ∠P. In △ABC, ∠C = 180° − ∠A − ∠B = 180° − 80° − 60° = 40°.
- Therefore ∠P = 40°.
A girl 90 cm tall walks away from a lamp-post 3.6 m high at 1.2 m/s. Find the length of her shadow after 4 seconds.
- Let AB be the lamp-post (3.6 m) and CD the girl (0.9 m), standing vertically. DE is her shadow of length x. After 4 s she has walked BD = 1.2 × 4 = 4.8 m.
- In △ABE and △CDE: ∠B = ∠D = 90° (both vertical to the ground) and ∠E is common. So by AA, △ABE ∼ △CDE.
- Corresponding sides are proportional: BE/DE = AB/CD, i.e. (4.8 + x)/x = 3.6/0.9 = 4.
- So 4.8 + x = 4x → 3x = 4.8 → x = 1.6. The shadow is 1.6 m long.
Common Mistakes
Two triangles with equal corresponding sides ratios but you can write the similarity in any vertex order, like △ABC ∼ △EDF.
Once you've checked the three facts, the triangles ARE similar, so it feels like the names are just labels you can shuffle.
The ORDER encodes the correspondence. △ABC ∼ △DEF means A↔D, B↔E, C↔F. Writing △ABC ∼ △EDF claims A↔E, B↔D — a different (usually false) matching. Always list vertices so that equal angles line up.
For polygons, equal corresponding angles alone make them similar (or proportional sides alone do).
For TRIANGLES one condition does imply the other (that's the whole point of AA and SSS), so students carry the shortcut over to all polygons.
That shortcut is special to triangles. For general polygons you need BOTH: equal angles AND proportional sides. A square and a rhombus (proportional sides, unequal angles) and a square and a rectangle (equal angles, non-proportional sides) are each NOT similar.
The BPT gives AD/AB = AE/EC.
AD, AB, AE and EC are all in the figure, so any ratio of them looks plausible — and AD/DB looks almost the same as AD/AB.
The BPT divides each side into the SAME two pieces and matches piece-to-piece: AD/DB = AE/EC (top piece over bottom piece, on each side). A valid variant is AD/AB = AE/AC, but never mix a part of one side with the WHOLE of the other.
AAA similarity is the same shortcut as ASA/SSS congruence — three equal angles make the triangles congruent.
The criteria look like the congruence rules from Class IX, so 'three angles equal' feels as strong as 'three sides equal'.
Equal angles give the same SHAPE, not the same SIZE — that's similarity, not congruence. A tiny and a huge equilateral triangle have identical angles (60° each) but are obviously not congruent. AAA ⇒ similar; you need a side to pin down size for congruence.
Quick Check
A line DE meets AB at D and AC at E in △ABC and DE ∥ BC. If AD = 4, DB = 6 and AE = 6, what is EC?
In △PQR, points E on PQ and F on PR give PE = 4, EQ = 4.5, PF = 8, FR = 9. Is EF ∥ QR?
Which single extra fact, with ∠A = ∠D already known, lets you conclude △ABC ∼ △DEF by the AA criterion?
△ABC ∼ △DEF with AB/DE = 2/5. If BC = 4 cm, what is EF?
Practice Problems
Easy
In △ABC, DE ∥ BC with D on AB and E on AC. If AD = 2 cm, AB = 6 cm and AE = 3 cm, find AC.
Since DE ∥ BC, the BPT (part-to-whole form) gives AD/AB = AE/AC.
So 2/6 = 3/AC → AC = 3 × 6/2 = 9 cm.
State whether EF ∥ QR in △PQR, given PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm.
Compute the two ratios: PE/EQ = 3.9/3 = 1.3, and PF/FR = 3.6/2.4 = 1.5.
Since 1.3 ≠ 1.5, the sides are NOT divided in the same ratio, so by the converse of the BPT, EF is not parallel to QR.
Medium
ABCD is a trapezium with AB ∥ DC. E lies on AD and F on BC with EF ∥ AB. Show that AE/ED = BF/FC.
Join AC, meeting EF at G.
Given AB ∥ DC and EF ∥ AB, so EF ∥ DC (lines parallel to the same line are parallel).
In △ADC, EG ∥ DC, so by the BPT: AE/ED = AG/GC … (1)
In △CAB, GF ∥ AB, so by the BPT: CG/GA = CF/FB, i.e. AG/GC = BF/FC … (2)
From (1) and (2): AE/ED = BF/FC.
In △POQ and △SOR, PQ ∥ RS and the segments meet at O. If PQ = 5 cm, RS = 8 cm and OQ = 4 cm, find OR.
PQ ∥ RS gives ∠P = ∠S and ∠Q = ∠R (alternate angles), and ∠POQ = ∠SOR (vertically opposite). So △POQ ∼ △SOR by AA.
Corresponding sides: PQ/SR = OQ/OR.
So 5/8 = 4/OR → OR = 4 × 8/5 = 32/5 = 6.4 cm.
Challenge
D is a point on side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² = CB · CD.
Compare △ABC and △DAC.
∠ACB = ∠DCA (the same angle C, common to both triangles).
∠BAC = ∠ADC (given).
So by AA, △ABC ∼ △DAC. Be careful with the correspondence: A↔D, B↔A, C↔C, i.e. △BAC ∼ △ADC.
Matching sides: CA/CD = CB/CA (side opposite the common angle C over the corresponding side).
Cross-multiplying: CA × CA = CB × CD, that is CA² = CB · CD.
CM and RN are medians of △ABC and △PQR respectively, and △ABC ∼ △PQR. Prove that CM/RN = AB/PQ.
Since △ABC ∼ △PQR: AB/PQ = BC/QR = CA/RP … (1) and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2).
CM is the median to AB, so AM = ½AB; RN is the median to PQ, so PN = ½PQ. From (1), AB/PQ = CA/RP, and since AM = ½AB, PN = ½PQ, we get AM/PN = CA/RP, i.e. AM/PN = AC/PR.
In △AMC and △PNR: ∠A = ∠P (from (2)) and AM/PN = AC/PR. So by SAS similarity, △AMC ∼ △PNR.
Corresponding sides of these similar triangles give CM/RN = AC/RP. But from (1), AC/RP = AB/PQ.
Therefore CM/RN = AB/PQ — corresponding medians are in the same ratio as corresponding sides.
Summary
You should now be able to explain:
- Similar figures have the same shape but not necessarily the same size. All congruent figures are similar, but not the reverse.
- Two polygons (same number of sides) are similar only if both hold: corresponding angles equal and corresponding sides proportional. The common ratio is the scale factor.
- Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides in the same ratio — AD/DB = AE/EC. Proved using equal-area triangles on the same base between the same parallels.
- Converse of the BPT: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side — your tool for proving lines parallel.
- Similarity criteria for triangles: AAA/AA (equal angles ⇒ proportional sides), SSS (proportional sides ⇒ equal angles), SAS (one equal angle between two proportional sides). Each needs only three facts.
- Write similarity with the correct vertex order: △ABC ∼ △DEF means A↔D, B↔E, C↔F.
- Similar triangles power indirect measurement — heights, distances, shadows.
What’s Next
Up to now you’ve reasoned about shapes using lengths, ratios and parallels — pure geometry, no grid. Next, in Coordinate Geometry, you’ll place points on the x–y plane and turn geometry into algebra: a distance formula to find how far apart two points are, and a section formula to find the point that divides a segment in a given ratio — the same “dividing in a ratio” idea you just met in the BPT, now with exact coordinates.