Chapter 14 — Similar Triangles

Two figures are congruent if they are identical in shape and size, so that one can be placed exactly on the other. Two figures are similar if they have the same shape but may differ in size. Every congruent pair is similar, but not the reverse. Two circles of any radii are similar; two squares are similar; two equilateral triangles are similar. Two rectangles need not be — a 2 × 3 rectangle and a 2 × 5 rectangle have the same angles but different proportions. So equal angles alone do not make polygons similar; the sides must also be in proportion.

For two polygons with the same number of sides, similarity requires two conditions together: (i) corresponding angles are equal, and (ii) corresponding sides are in the same ratio. This common ratio is the scale factor. A photograph enlarged from 4 cm × 6 cm to 8 cm × 12 cm has scale factor 2: every length doubles, every angle stays the same. A map at scale 1 : 50000 is similar to the land it shows with scale factor 1/50000.

Both conditions are needed for polygons in general. A square and a rectangle have all angles equal but sides not in proportion — not similar. A square and a rhombus have all sides in the same ratio but angles unequal — not similar. Similar polygons are written with corresponding vertices in the same order: quadrilateral ABCD ~ quadrilateral PQRS means A ↔ P, B ↔ Q, C ↔ R, D ↔ S, and AB/PQ = BC/QR = CD/RS = DA/SP.

For triangles a remarkable simplification occurs: if the angles are equal, the sides are automatically in proportion, and if the sides are in proportion, the angles are automatically equal. Either condition alone guarantees similarity. This is what makes triangles the basic tool of similarity, and the rest of the chapter proves why.

Everyday similarity: a scale model of a car, the shadow of a pole compared with the shadow of a metre stick, an image on a screen and the slide in the projector, the nested triangles formed by a ladder and the wall at different heights, the triangles formed by the eye, a coin held at arm's length, and the moon. Each is a case of the same shape at a different size, and each can be measured through the ratios that similarity preserves.

Notation: we write ΔABC ~ ΔDEF for similarity and ΔABC ≅ ΔDEF for congruence. The symbol ~ is read "is similar to". The order of letters carries the correspondence and must be respected: ΔABC ~ ΔDEF says angle A = angle D, angle B = angle E, angle C = angle F, and AB/DE = BC/EF = CA/FD.

Theorem. 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: In ΔABC, a line DE parallel to BC meets AB at D and AC at E. To prove: AD/DB = AE/EC.

Construction: Join BE and CD. Draw EL perpendicular to AB and DM perpendicular to AC.

Proof. Area of ΔADE = ½ × AD × EL, and area of ΔBDE = ½ × DB × EL. So area(ADE)/area(BDE) = AD/DB … (1). Similarly, area of ΔADE = ½ × AE × DM and area of ΔCDE = ½ × EC × DM, so area(ADE)/area(CDE) = AE/EC … (2). Now ΔBDE and ΔCDE stand on the same base DE and lie between the same parallels DE and BC, so they have equal areas: area(BDE) = area(CDE) … (3). From (1), (2) and (3), AD/DB = AE/EC. Proved.

The theorem is attributed to Thales of Miletus, who is said to have used it to measure the height of a pyramid from its shadow. It is the foundation of everything in this chapter.

Other forms of the ratio. From AD/DB = AE/EC, adding 1 to both sides gives (AD + DB)/DB = (AE + EC)/EC, that is, AB/DB = AC/EC; and taking reciprocals and adding 1 gives AB/AD = AC/AE. All three forms are used, and the examination accepts any.

Example 1. In ΔABC, DE ∥ BC, AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Find EC. AD/DB = AE/EC gives 1.5/3 = 1/EC, so EC = 2 cm.

Example 2. DE ∥ BC, AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x. x/(x − 2) = (x + 2)/(x − 1), so x(x − 1) = (x + 2)(x − 2), x² − x = x² − 4, x = 4.

Example 3. In ΔPQR, ST ∥ QR with PS = 3, SQ = 4.5, PT = 2. Find TR and PR. 3/4.5 = 2/TR, TR = 3, PR = 5.

Example 4 — a trapezium. In trapezium ABCD with AB ∥ DC, the diagonals meet at O. Show that AO/BO = CO/DO. Draw EF through O parallel to AB (and so to DC). In ΔADC, EO ∥ DC gives AE/ED = AO/OC; in ΔABD, EO ∥ AB gives AE/ED = BO/OD. So AO/OC = BO/OD, that is, AO/BO = CO/DO.

In a problem, identify the triangle, the parallel line, and which segments are "upper" and "lower" on each side; then write the proportion with the same orientation on both sides. The commonest error is to mix AD/DB with AE/AC.

Theorem (converse). If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given: In ΔABC, D on AB and E on AC with AD/DB = AE/EC. To prove: DE ∥ BC.

Proof (by contradiction). Suppose DE is not parallel to BC. Then draw DE' parallel to BC with E' on AC. By the Basic Proportionality Theorem, AD/DB = AE'/E'C. But we are given AD/DB = AE/EC. Hence AE/EC = AE'/E'C. Adding 1 to both sides: (AE + EC)/EC = (AE' + E'C)/E'C, so AC/EC = AC/E'C, so EC = E'C. Thus E and E' coincide, and DE is the same line as DE', which is parallel to BC. So DE ∥ BC. Proved.

The converse turns the theorem into a test for parallelism: measure the four segments, form the two ratios, and if they are equal the line is parallel to the third side.

Example 1. In ΔABC, D and E are on AB and AC with AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, EC = 5.5 cm. Is DE ∥ BC? AD/DB = 5.7/9.5 = 0.6 and AE/EC = 3.3/5.5 = 0.6. Equal, so yes.

Example 2. AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, AE = 1.8 cm. DB = 4.2, EC = 5.4. AD/DB = 1.4/4.2 = 1/3, AE/EC = 1.8/5.4 = 1/3. So DE ∥ BC. (Equivalently AD/AB = 1.4/5.6 = 1/4 = AE/AC = 1.8/7.2.)

Example 3. AD = 4, DB = 4.5, AE = 8, EC = 9. AD/DB = 8/9 = AE/EC. Parallel. Whereas AD = 4, DB = 4.5, AE = 8, EC = 10 gives 8/9 ≠ 4/5. Not parallel.

Application: the midpoint theorem. If D and E are the midpoints of AB and AC, then AD/DB = 1 = AE/EC, so by the converse DE ∥ BC. This is half of the midpoint theorem of Class 9; the other half, DE = ½ BC, follows from similarity later in this chapter.

Application: a quadrilateral's midpoints. Join the midpoints P, Q, R, S of the sides AB, BC, CD, DA of a quadrilateral. In ΔABC, PQ ∥ AC; in ΔADC, SR ∥ AC. So PQ ∥ SR; similarly QR ∥ PS, and PQRS is a parallelogram.

Application: a ratio inside a triangle. In ΔABC, D on AB and E on AC with AD/AB = AE/AC = 1/4; then AD/DB = 1/3 = AE/EC, so DE ∥ BC. This is the same configuration whose areas were compared in coordinate geometry, and here the parallelism is established without coordinates.

Proofs by contradiction of this kind — assume the opposite, derive a coincidence, conclude — are used again for the converse of the Pythagoras theorem. Learn the shape of the argument once.

The proportionality theorem underlies two constructions the examination sets regularly.

Construction 1: divide a segment AB in the ratio m : n (say 3 : 2). Draw a ray AX making an acute angle with AB. On AX mark m + n = 5 equal points A₁, A₂, …, A₅ with compasses. Join A₅ to B. Through A₃ (the mth point) draw a line parallel to A₅B, meeting AB at C. Then AC : CB = 3 : 2. Justification: in ΔAA₅B, A₃C ∥ A₅B, so by the Basic Proportionality Theorem AA₃/A₃A₅ = AC/CB, and AA₃/A₃A₅ = 3/2 by construction.

Construction 2: a triangle similar to a given triangle with sides 3/4 of the original (scale factor less than 1). Given ΔABC. Draw a ray BX below BC making an acute angle. Mark 4 equal points B₁, …, B₄ on BX (4 being the larger of 3 and 4). Join B₄C. Through B₃ draw a line parallel to B₄C meeting BC at C'. Through C' draw a line parallel to CA meeting BA at A'. Then ΔA'BC' ~ ΔABC with scale factor 3/4. Justification: B₃C' ∥ B₄C gives BC'/BC = BB₃/BB₄ = 3/4; C'A' ∥ CA gives BA'/BA = BC'/BC = 3/4 and the angles are equal, so the triangles are similar with sides in the ratio 3/4.

Construction 3: scale factor greater than 1, say 5/3. Mark 5 points on BX (the larger of 5 and 3). Join B₃ to C. Through B₅ draw a line parallel to B₃C meeting BC produced at C'. Through C' draw a line parallel to CA meeting BA produced at A'. Then ΔA'BC' ~ ΔABC with BC'/BC = 5/3. The new triangle contains the old one.

Points to present. Use a ruler and compasses only; keep the equal steps on the ray truly equal (same compass opening); draw the parallel with a set square or by copying the angle; label everything; write the two-line justification. In the SSC paper, a construction carries four marks: two for the figure, two for the justification.

Why the number of points is the larger of m and n. For a scale factor m/n, the mth point is joined to C when m < n (reduction) and the parallel is drawn through the nth point; when m > n (enlargement) the nth point is joined to C and the parallel is drawn through the mth. In both cases the number of points marked is max(m, n), so that both the joined point and the parallel point exist.

A related problem. To find a point P on AB such that AP = 2/5 AB, divide AB in the ratio 2 : 3 by Construction 1. To draw a line through a point D on AB parallel to BC without a set square, mark E on AC with AE/EC = AD/DB using the same construction, and join DE; by the converse theorem DE ∥ BC.

For triangles, similarity can be established from fewer facts than the definition requires. Three criteria are proved in this chapter.

AAA criterion. If in two triangles the corresponding angles are equal, then the corresponding sides are in the same ratio, and the triangles are similar. Proof sketch: Given ΔABC and ΔDEF with ∠A = ∠D, ∠B = ∠E, ∠C = ∠F. Cut DP = AB on DE and DQ = AC on DF and join PQ. Then ΔABC ≅ ΔDPQ (SAS), so ∠DPQ = ∠B = ∠E, hence PQ ∥ EF, hence by the BPT DP/PE = DQ/QF, which gives AB/DE = AC/DF. Similarly AB/DE = BC/EF.

AA corollary. Since the angles of a triangle sum to 180°, two equal pairs force the third; so two equal angles suffice. This is the criterion used most in practice.

SSS criterion. If the corresponding sides of two triangles are in the same ratio, then their corresponding angles are equal and the triangles are similar. Proof sketch: With AB/DE = BC/EF = CA/FD, cut DP = AB and DQ = AC and join PQ; then DP/PE = DQ/QF, so PQ ∥ EF, so ΔDPQ ~ ΔDEF and PQ = BC, giving ΔABC ≅ ΔDPQ and the angles equal.

SAS criterion. If one angle of a triangle equals one angle of another and the sides including these angles are in the same ratio, the triangles are similar. Proof sketch: With ∠A = ∠D and AB/DE = AC/DF, cut DP = AB, DQ = AC; ΔABC ≅ ΔDPQ (SAS), PQ ∥ EF by the converse BPT, so ∠DPQ = ∠E, and the AA criterion finishes.

Using the criteria. Always state the criterion and the correspondence. "In ΔABC and ΔPQR, ∠A = ∠P (given), ∠B = ∠Q (alternate angles); so ΔABC ~ ΔPQR by AA."

Example 1. ΔABC with sides 2.5, 3, 4 and ΔDEF with sides 5, 6, 8: ratios 2.5/5 = 3/6 = 4/8 = 1/2, so similar by SSS, with ∠A = ∠D etc. by matching the sides in size order.

Example 2. In ΔABC and ΔDEF, ∠A = 60°, ∠B = 80°, ∠D = 60°, ∠F = 40°. Then ∠E = 80°, so ∠A = ∠D and ∠B = ∠E: similar by AA. Note the correspondence is A ↔ D, B ↔ E, C ↔ F.

Example 3. In ΔPQR and ΔXYZ, PQ/XY = PR/XZ = 2/3 and ∠P = ∠X: similar by SAS.

Not similar. Sides 3, 4, 6 and 6, 8, 10: ratios 1/2, 1/2, 3/5 — not all equal, so not similar by SSS. Angle 50° included between sides 4 and 6, and angle 50° included between 6 and 8: ratios 2/3 and 3/4 unequal — SAS fails.

Common configurations that produce similar triangles. Two triangles with a common angle and a pair of parallel sides (the "nested" triangles ΔADE and ΔABC when DE ∥ BC). Two triangles on either side of a crossing of two lines with a pair of parallel sides (the "bow-tie", where the vertical angles are equal and alternate angles supply the second pair). The two small triangles cut off from a right triangle by the altitude to the hypotenuse, each similar to the whole. Recognising these three shapes on sight is most of the skill.

Examination questions ask for proofs of similarity in specific figures, often followed by a consequence about lengths. The three figures below cover most cases.

Figure 1: parallel lines cut by a transversal — the trapezium and its diagonals. In trapezium ABCD with AB ∥ DC, the diagonals meet at O. Prove ΔAOB ~ ΔCOD. ∠AOB = ∠COD (vertically opposite); ∠OAB = ∠OCD (alternate angles, AB ∥ DC). So ΔAOB ~ ΔCOD by AA, and AO/CO = BO/DO = AB/CD. If AB = 2CD, then AO = 2CO and BO = 2DO; and the areas are in the ratio 4 : 1.

Figure 2: a line through the midpoint. In ΔABC, D is the midpoint of AB and DE ∥ BC meets AC at E. Prove that E is the midpoint of AC and DE = ½ BC. ΔADE ~ ΔABC by AA (common angle A, corresponding angles from the parallel), so AE/AC = AD/AB = 1/2 and DE/BC = 1/2. Both parts of the midpoint theorem follow.

Figure 3: two right triangles sharing an acute angle. In ΔABC right-angled at B, BD ⊥ AC meets AC at D. Prove ΔADB ~ ΔBDC ~ ΔABC. ΔADB and ΔABC: ∠A common, ∠ADB = ∠ABC = 90°, so similar by AA. ΔBDC and ΔABC: ∠C common, ∠BDC = ∠ABC = 90°, similar by AA. Hence all three are similar. From ΔADB ~ ΔBDC, AD/BD = BD/DC, so BD² = AD × DC; from ΔADB ~ ΔABC, AB/AC = AD/AB, so AB² = AD × AC; and similarly BC² = CD × AC. These are the key results used later to prove Pythagoras.

Figure 4: the bisector of an angle. Although the angle bisector theorem is proved in the next course, a common problem gives ∠BAD = ∠CAE in two triangles with a shared vertex and asks for similarity; supply the second equal angle from the figure and apply AA.

Figure 5: two triangles with a common vertex and a shared angle. If in ΔABC, points P on AB and Q on AC satisfy AP × AB = AQ × AC, then AP/AC = AQ/AB, and with the common angle A, ΔAPQ ~ ΔACB by SAS (note the reversed correspondence: P ↔ C, Q ↔ B). Hence ∠APQ = ∠ACB, and PQ is an "antiparallel" to BC.

Figure 6: a chord-like crossing. Two segments AB and CD meet at P so that PA × PB = PC × PD. Then PA/PC = PD/PB and ∠APC = ∠DPB (vertically opposite), so ΔAPC ~ ΔDPB by SAS, giving ∠PAC = ∠PDB.

Writing the proof. Name the two triangles with corresponding vertices in order. List the equal angles or the proportional sides with a reason for each. State the criterion. Then draw the consequence for the sides or areas that the question asks about. Two lines for the facts, one for the criterion, one for the conclusion: a four-line proof is complete.

Because the sun's rays are parallel, the shadow of every vertical object at a given moment makes the same angle with the ground, so the triangles formed by objects and their shadows are all similar. Because light travels in straight lines, images and objects form similar triangles with a mirror or a pinhole. These facts let us measure what we cannot reach.

Example 1 — a tower from a stick. A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Find the height of the tower. The two right triangles have the same angle of elevation of the sun, so they are similar by AA. height/shadow is the same: h/30 = 10/8, so h = 300/8 = 37.5 m.

Example 2 — a girl and a lamp post. A girl of height 90 cm walks away from the base of a lamp post at 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. After 4 s she is 4.8 m from the post. Let her shadow be x m. The triangle formed by the lamp top, the post base and the shadow tip is similar to the triangle formed by her head, her feet and the shadow tip (both right triangles sharing the angle at the shadow tip). So 3.6/(4.8 + x) = 0.9/x, giving 3.6x = 0.9(4.8 + x), 3.6x = 4.32 + 0.9x, 2.7x = 4.32, x = 1.6 m.

Example 3 — a mirror on the ground. A boy 1.5 m tall stands 3 m from a mirror lying on the ground and sees the top of a building whose base is 24 m from the mirror. Since the angle of incidence equals the angle of reflection, the two right triangles are similar: h/24 = 1.5/3, h = 12 m.

Example 4 — the width of a river. To find the width of a river, a surveyor marks A on the near bank opposite a tree T on the far bank, walks 40 m along the bank to B, then 10 m further to C, and then walks inland perpendicular to the bank to a point D from which T, B and D are in line; CD = 15 m. Triangles ABT and CBD are similar (right angles at A and C, vertically opposite angles at B), so AT/CD = AB/BC, AT = 15 × 40/10 = 60 m.

Example 5 — a ladder. A ladder rests against a wall; a rung 2 m along the ladder from the foot is 1.2 m above the ground. If the ladder is 5 m long, how high up the wall does it reach? The small triangle (foot, point below the rung, rung) and the whole triangle (foot, wall base, top) are similar: h/5 = 1.2/2, h = 3 m.

Example 6 — nested triangles on a hillside. Two poles of heights 6 m and 11 m stand 12 m apart; a wire from the top of the shorter pole to the top of the taller makes a right triangle with legs 12 and 5, so the wire is 13 m; a point on the wire 2.6 m from the shorter pole is (by similarity, 2.6/13 = 1/5) 1 m higher than the shorter pole, at height 7 m.

In each case: draw the two triangles, mark the right angles and the common angle, state the similarity, write the ratio of corresponding sides, and solve. Convert all lengths to the same unit before computing.

Theorem. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ΔABC ~ ΔPQR. To prove: area(ABC)/area(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)².

Construction: Draw AM ⊥ BC and PN ⊥ QR.

Proof. area(ABC)/area(PQR) = (½ × BC × AM)/(½ × QR × PN) = (BC/QR) × (AM/PN) … (1). Now in ΔABM and ΔPQN, ∠B = ∠Q (similar triangles) and ∠M = ∠N = 90°, so ΔABM ~ ΔPQN by AA, giving AM/PN = AB/PQ. But AB/PQ = BC/QR since the triangles are similar. So AM/PN = BC/QR … (2). From (1) and (2), area(ABC)/area(PQR) = (BC/QR) × (BC/QR) = (BC/QR)², and the same for the other pairs of sides. Proved.

The theorem also gives the ratio in terms of corresponding altitudes, medians or angle bisectors, since those are all in the same ratio as the sides: area ratio = (altitude ratio)² = (median ratio)².

Example 1. ΔABC ~ ΔDEF with BC = 3 cm, EF = 4 cm and area(ABC) = 54 cm². Area(DEF) = 54 × (4/3)² = 54 × 16/9 = 96 cm².

Example 2. The areas of two similar triangles are 81 cm² and 49 cm². If the altitude of the first is 4.5 cm, find the altitude of the second. (4.5/h)² = 81/49, so 4.5/h = 9/7, h = 3.5 cm.

Example 3. D, E, F are the midpoints of the sides of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC. Each side of DEF is half the corresponding side of ABC (midpoint theorem), and ΔDEF ~ ΔABC, so the ratio is (1/2)² = 1/4.

Example 4. In the trapezium ABCD with AB ∥ DC and AB = 2CD, diagonals meet at O. ΔAOB ~ ΔCOD with AB/CD = 2, so area(AOB) : area(COD) = 4 : 1.

Example 5. In ΔABC, DE ∥ BC with AD/DB = 3/2. Find area(ADE)/area(trapezium DBCE). ΔADE ~ ΔABC with AD/AB = 3/5, so area(ADE)/area(ABC) = 9/25, hence area(DBCE) = 16/25 of ABC, and the required ratio is 9 : 16.

Example 6. Two similar triangles have equal areas. Prove they are congruent. The area ratio is 1, so the side ratio is 1, so all corresponding sides are equal: congruent by SSS.

Example 7. Triangles ABC and PQR are similar with areas 100 cm² and 144 cm². If PQ = 12 cm, find AB. (AB/12)² = 100/144, AB/12 = 10/12, AB = 10 cm.

Do not confuse the two ratios: sides go as k, areas as k². A triangle with sides doubled has four times the area, with sides tripled nine times. When the areas are given and a side is asked, take the square root of the area ratio.

Theorem (Pythagoras). In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: ΔABC right-angled at B. To prove: AC² = AB² + BC².

Construction: Draw BD ⊥ AC, meeting AC at D.

Proof. In ΔADB and ΔABC: ∠A is common and ∠ADB = ∠ABC = 90°, so ΔADB ~ ΔABC (AA). Hence AD/AB = AB/AC, so AB² = AD × AC … (1). In ΔBDC and ΔABC: ∠C is common and ∠BDC = ∠ABC = 90°, so ΔBDC ~ ΔABC (AA). Hence CD/BC = BC/AC, so BC² = CD × AC … (2). Adding (1) and (2): AB² + BC² = AD × AC + CD × AC = AC(AD + CD) = AC × AC = AC². Proved.

This proof, using the altitude to the hypotenuse and the two similar triangles it creates, is the one the Telangana syllabus requires. The Class 9 chapter used areas; here similarity does the work.

The theorem in use. A ladder 10 m long reaches a window 8 m above the ground: its foot is √(100 − 64) = 6 m from the wall. A rectangle with sides 5 and 12 has diagonal 13. The diagonal of a square of side a is a√2. The altitude of an equilateral triangle of side a is (√3/2)a, since a² = (a/2)² + h² gives h² = 3a²/4.

Pythagorean triples. Sets of whole numbers satisfying a² + b² = c²: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), and any multiple of these such as (6, 8, 10), (9, 12, 15), (10, 24, 26). Recognising a triple saves computation.

Example 1. A man goes 15 m due west and then 8 m due north. How far is he from the start? √(225 + 64) = √289 = 17 m.

Example 2. Two poles of heights 6 m and 11 m stand on level ground 12 m apart. Find the distance between their tops. The difference of heights is 5 m and the horizontal gap is 12 m, so the distance is √(25 + 144) = 13 m.

Example 3. In an isosceles triangle with equal sides 13 cm and base 10 cm, find the altitude to the base. It bisects the base: h = √(169 − 25) = 12 cm. Area = ½ × 10 × 12 = 60 cm².

Example 4 — the three results from the altitude. In the figure of the proof, BD² = AD × DC as well (from ΔADB ~ ΔBDC). If AD = 4 and DC = 9, then BD = 6, AB² = 4 × 13 = 52, BC² = 9 × 13 = 117, and AB² + BC² = 169 = AC². Consistent.

Example 5. Prove that in a right triangle the altitude to the hypotenuse satisfies 1/BD² = 1/AB² + 1/BC². From area: AB × BC = AC × BD, so BD = AB·BC/AC, so 1/BD² = AC²/(AB²·BC²) = (AB² + BC²)/(AB²·BC²) = 1/BC² + 1/AB².

Theorem (converse). In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Given: ΔABC with AC² = AB² + BC². To prove: ∠B = 90°.

Construction: Draw ΔPQR right-angled at Q with PQ = AB and QR = BC.

Proof. In ΔPQR, by Pythagoras, PR² = PQ² + QR² = AB² + BC² = AC² (given). So PR = AC. Now in ΔABC and ΔPQR: AB = PQ, BC = QR, AC = PR, so ΔABC ≅ ΔPQR by SSS. Hence ∠B = ∠Q = 90°. Proved.

The converse gives a test for a right angle from three lengths, without a protractor. It is used by builders who lay out a right angle with a 3-4-5 rope, and it appears in the examination as "determine whether the triangle with the given sides is right-angled".

Example 1. Sides 7, 24, 25: 7² + 24² = 49 + 576 = 625 = 25². Right-angled, with the right angle opposite the 25 side.

Example 2. Sides 3, 8, 6: the longest is 8; 3² + 6² = 45 ≠ 64. Not right-angled (in fact obtuse, since 45 < 64).

Example 3. Sides 50, 80, 100: 2500 + 6400 = 8900 ≠ 10000. Not right-angled.

Example 4. Sides 13, 12, 5: 25 + 144 = 169 = 13². Right-angled at the vertex opposite 13.

Extension (not for proof, useful for insight). If c is the longest side: c² < a² + b² means the angle opposite c is acute; c² > a² + b² means it is obtuse. So 3, 8, 6 is obtuse and 5, 6, 7 (49 < 61) is acute.

Example 5 — an application. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ∠C is a right angle. AB² = 2AC² = AC² + AC² = AC² + BC². So by the converse, the angle opposite AB, which is ∠C, is 90°.

Example 6. In ΔABC, AB = 6√3, AC = 12, BC = 6. Is it right-angled? AB² = 108, AC² = 144, BC² = 36; 108 + 36 = 144. Yes, right-angled at B, and since BC = ½ AC, ∠A = 30°.

Example 7 — the rhombus. Prove that the sum of the squares of the sides of a rhombus equals the sum of the squares of its diagonals. The diagonals bisect each other at right angles at O. In right ΔAOB, AB² = OA² + OB² = (AC/2)² + (BD/2)². So 4AB² = AC² + BD², and since all four sides equal AB, AB² + BC² + CD² + DA² = AC² + BD².

When testing, always identify the longest side first and compare its square with the sum of the other two squares; testing the wrong side gives a false "not right-angled".

The chapter's later exercises combine Pythagoras with similarity, medians and the altitude-to-hypotenuse results. Several standard proofs recur.

Problem 1. In ΔABC right-angled at C, D is the midpoint of BC. Prove that AB² = 4AD² − 3AC². AD² = AC² + CD² = AC² + (BC/2)², so 4AD² = 4AC² + BC². And AB² = AC² + BC² = AC² + (4AD² − 4AC²) = 4AD² − 3AC².

Problem 2. In ΔABC right-angled at C, D and E are on CA and CB. Prove AE² + BD² = AB² + DE². AE² = AC² + CE², BD² = BC² + CD². Adding: AE² + BD² = (AC² + BC²) + (CE² + CD²) = AB² + DE².

Problem 3. The perpendicular from A on side BC of ΔABC meets BC at D such that DB = 3CD. Prove 2AB² = 2AC² + BC². Let CD = x, so DB = 3x and BC = 4x. AB² = AD² + 9x², AC² = AD² + x². So AB² − AC² = 8x² = ½ (4x)² = ½ BC², giving 2AB² = 2AC² + BC².

Problem 4. In an equilateral triangle ABC, D is a point on BC with BD = ⅓ BC. Prove 9AD² = 7AB². Let AB = a and E be the midpoint of BC, so AE = (√3/2)a, BE = a/2, BD = a/3, DE = a/2 − a/3 = a/6. AD² = AE² + DE² = 3a²/4 + a²/36 = (27a² + a²)/36 = 28a²/36 = 7a²/9. So 9AD² = 7a² = 7AB².

Problem 5. Prove that three times the square of any side of an equilateral triangle equals four times the square of the altitude. With side a and altitude h: h² = a² − a²/4 = 3a²/4, so 4h² = 3a².

Problem 6 — an aeroplane problem. An aeroplane leaves an airport and flies due north at 1000 km/h. At the same time another leaves the same airport and flies due west at 1200 km/h. How far apart are they after 1½ hours? Distances 1500 km and 1800 km at right angles: √(1500² + 1800²) = √(2250000 + 3240000) = √5490000 = 300√61 km ≈ 2343 km.

Problem 7 — a wire from a pole. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake at the other end. How far is the stake from the base? √(576 − 324) = √252 = 6√7 m ≈ 15.9 m.

Problem 8 — a ladder moved. A 5 m ladder has its foot 3 m from a wall and its top 4 m up. If the foot is moved to 4 m from the wall, the top comes down to 3 m: it slides 1 m.

Problem 9 — O inside a rectangle. O is any point inside rectangle ABCD. Prove OB² + OD² = OA² + OC². Draw through O a line parallel to AB meeting AD at P and BC at Q. Then OB² + OD² = (OQ² + QB²) + (OP² + PD²) and OA² + OC² = (OP² + PA²) + (OQ² + QC²); since QB = PA and PD = QC, the sums are equal.

The pattern in every proof: express each squared length by Pythagoras in a right triangle of the figure, then add or subtract to reach the target. Introduce a letter for a segment when the ratio is given, as in Problem 3, and use midpoints and altitudes to create right angles.

The chapter contains five theorems whose proofs are examinable, and a body of applications.

The theorems. (1) Basic Proportionality Theorem: DE ∥ BC ⇒ AD/DB = AE/EC, proved by areas. (2) Its converse, proved by contradiction. (3) Areas of similar triangles are as the squares of corresponding sides, proved via similar right triangles on the altitudes. (4) Pythagoras, proved by the altitude to the hypotenuse and two similarities. (5) Converse of Pythagoras, proved by constructing a right triangle and using SSS congruence. The three similarity criteria AA, SSS, SAS are stated and their proofs sketched; the examination asks for their use, not their proof.

One-mark questions ask whether two given triangles are similar and by which criterion; the ratio of areas from a ratio of sides; whether given lengths form a right triangle; a missing length by the BPT.

Two-mark questions ask for a length from the BPT with an unknown x; the test DE ∥ BC from four lengths; a side or altitude from an area ratio; a distance by Pythagoras (ladder, poles, journeys); the altitude of an equilateral triangle.

Four-mark questions ask for one of the five proofs; a construction (dividing a segment, a similar triangle with a given scale factor) with justification; a similarity proof in a figure followed by a consequence (BD² = AD × DC, AO/CO = BO/DO, DE = ½ BC); a Pythagoras-based proof (AB² = 4AD² − 3AC², the rhombus identity, 9AD² = 7AB²); a practical problem (shadow, mirror, river).

Proof presentation. Given, To prove, Construction (if any), Proof with numbered steps and a reason for each, and the conclusion restated. A figure is compulsory and must match the statement. The examination allots marks for the figure, for the construction, for the correct use of the criterion, and for the conclusion.

Errors to avoid. Writing the correspondence in the wrong order (ΔABC ~ ΔDEF when A actually corresponds to E). Confusing the ratio of sides with the ratio of areas. Applying the BPT to a line that is not parallel to a side. Testing the converse of Pythagoras with the wrong side as hypotenuse. Forgetting that AA needs the equal angles to be corresponding. In constructions, marking too few points on the ray or drawing the parallel through the wrong point. In shadow problems, mixing metres and centimetres.

Where this leads. Trigonometry, the very next chapters, defines sine, cosine and tangent as ratios of sides in a right triangle — ratios that are the same for all similar right triangles, which is precisely why the definitions make sense. The Intermediate course continues with the angle bisector theorem, the geometry of circles, and vectors, all resting on similarity.