Chapter 15 — Similar Triangles Optional

The main exercises give a figure with the similar triangles already visible and ask for a length or a one-step proof. The optional exercise gives a figure in which the similar triangles must be found, often after a construction, and asks for a relation between products or squares of lengths that follows from two or three steps. The theorems used are the same — the Basic Proportionality Theorem and its converse, the AA, SSS and SAS criteria, the area theorem, and Pythagoras with its converse — but the work of the problem is in seeing which pair of triangles to compare.

Three shapes account for almost every similarity in this exercise. The nested shape: a small triangle inside a large one, sharing a vertex, with a side parallel to a side of the large one (ΔADE inside ΔABC with DE ∥ BC). The bow-tie: two triangles meeting at a point with vertically opposite angles and a pair of parallel sides (ΔAOB and ΔCOD across the intersection O of the diagonals of a trapezium). The altitude-to-hypotenuse shape: a right triangle split by the perpendicular from the right angle into two triangles similar to it and to each other. Whenever a product of two segments equals a product of two others, look for one of these shapes; the equality of products is a proportion in disguise.

The Pythagoras problems have a different rhythm. They give a relation between squares — AB² = 4AD² − 3AC², 9AD² = 7AB², 2AB² = 2AC² + BC² — and ask for a proof. The method is always: find the right triangles in the figure (create one with an altitude or a midpoint if needed), write Pythagoras for each, and add or subtract to eliminate the unwanted segments. Introduce a letter for a segment length when a ratio is given.

Presentation. Every proof begins with the figure and the three lines Given, To prove, Construction. Each similarity is stated with its criterion and the correspondence in order. Each proportion is written from the similarity, and the target is reached by cross-multiplying, adding or substituting. The final line restates what was to be proved. Four marks are awarded as: figure and setup, the key similarity or Pythagoras statements, the algebra, and the conclusion. A correct final line without the similarity named earns little.

The chapter now takes the families in order: product relations from similar triangles; the trapezium and its diagonals; the altitude to the hypotenuse; identities among squares; areas of similar triangles in composite figures; the rhombus, the rectangle and other quadrilaterals; and numerical problems on wires, ladders and moving objects.

An equation of the form PQ × RS = TU × VW is the cross-multiplied form of a proportion PQ/TU = VW/RS, and a proportion comes from a pair of similar triangles. To prove such a relation, find two triangles containing the four segments as corresponding sides, prove them similar, and cross-multiply.

Example 1. In the figure, ∠ABC = 90° and BD ⊥ AC. Prove that BD² = AD × DC. In ΔADB and ΔBDC: ∠ADB = ∠BDC = 90°; ∠DAB = ∠DBC, because each is the complement of ∠ABD (∠DAB + ∠ABD = 90° in ΔADB, and ∠DBC + ∠ABD = 90° since ∠ABC = 90°). So ΔADB ~ ΔBDC by AA, with A ↔ B, D ↔ D, B ↔ C. Hence AD/BD = DB/DC, so BD² = AD × DC.

Example 2. D is a point on side BC of ΔABC such that ∠ADC = ∠BAC. Prove that CA² = CB × CD. In ΔADC and ΔBAC: ∠ADC = ∠BAC (given) and ∠C is common. So ΔADC ~ ΔBAC by AA, with A ↔ B, D ↔ A, C ↔ C. Hence CA/CB = CD/CA, so CA² = CB × CD.

Example 3. In ΔABC, AD ⊥ BC and AD² = BD × DC. Prove that ∠BAC = 90°. From AD² = BD × DC, AD/BD = DC/AD, that is, BD/AD = AD/DC. In ΔADB and ΔCDA, ∠ADB = ∠CDA = 90° and the sides about these angles are proportional (BD/AD = AD/DC), so ΔADB ~ ΔCDA by SAS with B ↔ A, D ↔ D, A ↔ C. Hence ∠BAD = ∠ACD and ∠ABD = ∠CAD. Then ∠BAC = ∠BAD + ∠CAD = ∠ACD + ∠ABD = 180° − ∠BAC (angle sum in ΔABC), so 2∠BAC = 180° and ∠BAC = 90°.

Example 4. Two chords AB and CD of a circle intersect at P inside the circle. Prove that PA × PB = PC × PD. Join AC and BD. In ΔAPC and ΔDPB: ∠APC = ∠DPB (vertically opposite) and ∠CAP = ∠BDP (angles in the same segment, subtending arc CB). So ΔAPC ~ ΔDPB by AA, and PA/PD = PC/PB, giving PA × PB = PC × PD. (This is the intersecting chords theorem, which the next chapter takes up.)

Example 5. E is a point on side AD produced of a parallelogram ABCD, and BE intersects CD at F. Prove that ΔABE ~ ΔCFB. ∠A = ∠C (opposite angles of a parallelogram); ∠AEB = ∠CBF (alternate angles, AE ∥ BC with BE as transversal). So ΔABE ~ ΔCFB by AA, with A ↔ C, B ↔ F, E ↔ B. Consequently AB/CF = AE/CB = BE/FB.

Example 6. In ΔABC, D and E are points on AB and AC with ∠ADE = ∠ACB. Prove that AD × AB = AE × AC. In ΔADE and ΔACB: ∠ADE = ∠ACB (given), ∠A common; so ΔADE ~ ΔACB by AA with D ↔ C, E ↔ B. Then AD/AC = AE/AB, so AD × AB = AE × AC. Note the correspondence is reversed from the nested case; DE is antiparallel, not parallel, to BC.

In every example the correspondence of vertices was written explicitly. That is what makes the proportion come out right; guessing the proportion from the letters without the correspondence is the commonest source of a wrong product.

A trapezium with its diagonals is the richest single figure of the chapter, containing both a bow-tie and nested triangles.

Result 1. In trapezium ABCD with AB ∥ DC, the diagonals AC and BD meet at O. Then AO/OC = BO/OD. Proof by similarity: ΔAOB ~ ΔCOD by AA (vertically opposite angles at O; alternate angles ∠OAB = ∠OCD), so AO/CO = BO/DO = AB/CD. Proof by BPT: draw EF through O parallel to AB (and DC), with E on AD and F on BC. In ΔADC, EO ∥ DC gives AE/ED = AO/OC. In ΔDAB, EO ∥ AB gives AE/ED = BO/OD. So AO/OC = BO/OD.

Result 2 (converse). If the diagonals of quadrilateral ABCD meet at O with AO/OC = BO/OD, then ABCD is a trapezium. Proof: draw OE ∥ AB meeting AD at E. In ΔDAB, OE ∥ AB gives AE/ED = BO/OD = AO/OC (given). So in ΔADC, AE/ED = AO/OC, and by the converse of the BPT, EO ∥ DC. Since EO ∥ AB as well, AB ∥ DC, so ABCD is a trapezium.

Result 3. If AB = 2CD, then area(AOB) : area(COD) = 4 : 1 (area theorem, ratio of sides 2). If the areas are given as 84 cm² and 21 cm², then (AB/CD)² = 4, AB = 2CD.

Result 4 — the length of the parallel through O. Let AB = a and DC = b, and let EF through O be parallel to them. From ΔAOB ~ ΔCOD, AO/AC = a/(a + b). In ΔADC, EO ∥ DC gives EO/DC = AO/AC, so EO = ab/(a + b). Similarly OF = ab/(a + b), so EF = 2ab/(a + b), the harmonic mean of the parallel sides. For a = 6 and b = 3, EF = 36/9 = 4.

Result 5 — a parallel through any point of a diagonal. In ΔABC with a line through a point P on AC parallel to BC meeting AB at Q: AQ/QB = AP/PC. Two parallels through two points give two ratios; the ratios multiply and divide as segments do.

Example — three parallel lines cut by two transversals. Lines l, m, n are parallel and are cut by transversals p and q at A, B, C and D, E, F respectively. Prove AB/BC = DE/EF. Join AF, meeting m at G. In ΔACF, BG ∥ CF gives AB/BC = AG/GF. In ΔAFD, GE ∥ AD gives AG/GF = DE/EF. So AB/BC = DE/EF. This is the intercept theorem, an extension of the BPT.

Example — a line through O parallel to the sides of a triangle. In ΔABC, points D on AB and E on AC with DE ∥ BC; O is the intersection of BE and CD. Prove that DO/OC = EO/OB. ΔDOE ~ ΔCOB (bow-tie: vertically opposite angles at O, alternate angles from DE ∥ BC), so DO/CO = EO/BO, which rearranges to the required.

Draw the trapezium with the longer parallel side at the bottom; mark O; draw the parallel EF when needed; and note which of the two proofs is being asked. The BPT proof is expected when the question says "using the Basic Proportionality Theorem".

In ΔABC right-angled at B, the perpendicular BD from B to the hypotenuse AC creates three similar triangles: ΔADB, ΔBDC and ΔABC. This configuration yields a family of results the optional exercise draws on repeatedly.

The similarities. ΔADB ~ ΔABC (∠A common, right angles at D and B). ΔBDC ~ ΔABC (∠C common, right angles at D and B). Hence ΔADB ~ ΔBDC, with A ↔ B, D ↔ D, B ↔ C.

The three product relations. From ΔADB ~ ΔABC: AD/AB = AB/AC, so AB² = AD × AC. From ΔBDC ~ ΔABC: DC/BC = BC/AC, so BC² = DC × AC. From ΔADB ~ ΔBDC: AD/BD = BD/DC, so BD² = AD × DC. Adding the first two gives Pythagoras.

The altitude and the area. Area = ½ AB × BC = ½ AC × BD, so BD = AB × BC / AC. For legs 6 and 8, hypotenuse 10, the altitude is 4.8.

The reciprocal relation. 1/BD² = AC²/(AB² × BC²) = (AB² + BC²)/(AB² × BC²) = 1/AB² + 1/BC².

Example 1. AD = 4 cm, DC = 9 cm. Find BD, AB, BC. BD² = 36, BD = 6. AB² = 4 × 13 = 52, AB = 2√13. BC² = 9 × 13 = 117, BC = 3√13. Check: 52 + 117 = 169 = 13².

Example 2. AB = 5, BC = 12. Find AD, DC, BD. AC = 13. AD = AB²/AC = 25/13, DC = 144/13, BD = 60/13.

Example 3. Prove that in a right triangle the ratio AD : DC = AB² : BC². AD = AB²/AC and DC = BC²/AC, so AD/DC = AB²/BC². For legs 3 and 4, the altitude divides the hypotenuse in the ratio 9 : 16.

Example 4 — the median to the hypotenuse. If M is the midpoint of AC, then BM = AM = MC = AC/2. Proof: complete the rectangle ABCE with diagonals AC and BE meeting at M; the diagonals of a rectangle are equal and bisect each other, so BM = ½ BE = ½ AC. Hence the circumcentre of a right triangle is the midpoint of the hypotenuse.

Example 5 — the converse via the altitude. If in ΔABC the foot D of the altitude from A satisfies AD² = BD × DC, then ∠A = 90° (proved in the product-relations topic by SAS similarity).

Example 6 — a numeric application. A ladder leaning against a wall touches a 1 m high box placed against the wall so that the ladder's foot is 2 m from the box; the wall is reached at height h. With the altitude configuration reversed — the box corner D on the ladder, the ladder foot F, the wall base W and the top T — ΔFDB (box) ~ ΔFTW, so h/3 = 1/2 if the foot is 3 m from the wall, giving h = 1.5 m.

Whenever a right triangle appears with its altitude to the hypotenuse, write the three product relations at once; the question is almost always one of them or a combination.

The Pythagoras problems of the optional exercise ask for a proof of an identity like AB² = 4AD² − 3AC². The method: locate the right triangles, write Pythagoras in each, use the midpoint or ratio to relate the segments, and eliminate.

Problem 1. ΔABC is right-angled at C, and D is the midpoint of BC. Prove AB² = 4AD² − 3AC². In right ΔACD: AD² = AC² + CD² = AC² + BC²/4, so 4AD² = 4AC² + BC². In right ΔACB: AB² = AC² + BC². Substitute BC² = 4AD² − 4AC²: AB² = AC² + 4AD² − 4AC² = 4AD² − 3AC².

Problem 2. ΔABC is right-angled at C, with D and E the midpoints of CA and CB. Prove 4(AE² + BD²) = 5AB². AE² = AC² + CE² = AC² + BC²/4; BD² = BC² + CD² = BC² + AC²/4. Adding: AE² + BD² = (5/4)(AC² + BC²) = (5/4)AB², so 4(AE² + BD²) = 5AB².

Problem 3. The perpendicular from A on BC meets BC at D such that DB = 3CD. Prove 2AB² = 2AC² + BC². Let CD = x; then DB = 3x and BC = 4x. AB² = AD² + 9x², AC² = AD² + x². Subtract: AB² − AC² = 8x². And BC² = 16x², so 8x² = ½ BC². Thus AB² − AC² = ½ BC², i.e. 2AB² = 2AC² + BC².

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

Problem 5. In ΔABC, AD ⊥ BC and D lies between B and C. Prove AB² − AC² = BD² − CD². AB² = AD² + BD² and AC² = AD² + CD²; subtracting, AB² − AC² = BD² − CD². (If D lies on BC produced the same identity holds.)

Problem 6 — Apollonius' theorem. In ΔABC, AD is the median to BC. Prove AB² + AC² = 2AD² + 2BD². Draw AE ⊥ BC. Then AB² = AE² + BE² = AE² + (BD + DE)² = AE² + BD² + 2BD·DE + DE² and AC² = AE² + CE² = AE² + (CD − DE)² = AE² + CD² − 2CD·DE + DE². Adding, and using BD = CD so the cross terms cancel: AB² + AC² = 2AE² + 2DE² + 2BD² = 2(AE² + DE²) + 2BD² = 2AD² + 2BD². This is the standard result for a median, and it gives the length of a median from the sides: 4AD² = 2AB² + 2AC² − BC².

Problem 7. In ΔABC with ∠B obtuse and AD ⊥ CB produced, prove AC² = AB² + BC² + 2BC·BD. AC² = AD² + DC² = AD² + (DB + BC)² = (AD² + DB²) + BC² + 2BC·BD = AB² + BC² + 2BC·BD. The analogous acute case gives AC² = AB² + BC² − 2BC·BD. These are the geometric forms of the cosine rule.

Always name the right angle whose Pythagoras you are writing; introduce a letter for the unit segment when a ratio like DB = 3CD is given; and check the identity numerically with a specific triangle once the proof is done.

The area theorem — areas of similar triangles are as the squares of corresponding sides — is applied in the optional exercise to figures where the similar triangles are parts of a larger figure, and the required ratio involves a trapezium or a difference of areas.

Example 1. In ΔABC, DE ∥ BC with AD/DB = 3/2. Find area(ADE) : area(DBCE). ΔADE ~ ΔABC with AD/AB = 3/5, so area(ADE)/area(ABC) = 9/25. Then area(DBCE) = area(ABC) − area(ADE) = (16/25) area(ABC), and the ratio is 9 : 16.

Example 2. DE ∥ BC and area(ADE) = area(DBCE). Find AD/AB and BD/AB. area(ADE) = ½ area(ABC), so (AD/AB)² = 1/2, AD/AB = 1/√2, and BD/AB = 1 − 1/√2 = (√2 − 1)/√2 = (2 − √2)/2.

Example 3. D, E, F are the midpoints of BC, CA, AB of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC. By the midpoint theorem EF = ½ BC, FD = ½ CA, DE = ½ AB, so ΔDEF ~ ΔABC (SSS) with ratio 1/2, and the area ratio is 1 : 4. The four small triangles AFE, FBD, EDC and DEF are all congruent, each one-quarter of ABC.

Example 4. Trapezium ABCD with AB ∥ DC, AB = 2CD, diagonals meeting at O. Find area(AOB) : area(COD). ΔAOB ~ ΔCOD with AB/CD = 2, so 4 : 1. Further, area(AOD) = area(BOC) (triangles ABD and ABC on the same base AB between the same parallels have equal areas; subtract the common ΔAOB). And since AO : OC = 2 : 1, area(AOD) : area(COD) = 2 : 1 (same height from D). So the four triangles have areas in the ratio AOB : BOC : COD : DOA = 4 : 2 : 1 : 2.

Example 5. ΔABC ~ ΔPQR with area(ABC) = 4 × area(PQR). If BC = 12 cm, find QR. (BC/QR)² = 4, BC/QR = 2, QR = 6 cm.

Example 6. Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights. Isosceles triangles with equal vertical angles have equal base angles too, so they are similar (AA), and heights are in the ratio √(16/25) = 4 : 5.

Example 7. The diagonal BD of parallelogram ABCD meets AE at F, where E is on DC. Prove that DF × AF = FB × EF. ΔAFB ~ ΔEFD (bow-tie: vertically opposite angles at F, alternate angles from AB ∥ DE), so AF/EF = FB/FD. Cross-multiplying, AF × FD = FB × EF, that is, DF × AF = FB × EF. Always derive the product from the proportion rather than guessing which segments pair up.

Example 8. ΔABC and ΔDBC are on the same base BC, and AD meets BC at O. Prove area(ABC)/area(DBC) = AO/DO. Draw AM ⊥ BC and DN ⊥ BC. ΔAMO ~ ΔDNO (right angles, vertically opposite angles at O), so AM/DN = AO/DO. Then area(ABC)/area(DBC) = (½ BC × AM)/(½ BC × DN) = AM/DN = AO/DO.

The general habit: identify the similar pair and its side ratio k, write the area ratio k², then handle the rest of the figure by subtraction or by "same base, same height" reasoning.

Several optional problems concern quadrilaterals whose diagonals or interior points create right triangles.

Problem 1 — the rhombus. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. The diagonals of a rhombus bisect each other at right angles; let them meet at O. In right ΔAOB: AB² = OA² + OB² = (AC/2)² + (BD/2)² = (AC² + BD²)/4. So 4AB² = AC² + BD², and since AB = BC = CD = DA, AB² + BC² + CD² + DA² = AC² + BD².

Problem 2 — a numerical rhombus. The diagonals of a rhombus are 16 cm and 12 cm. Find its side and area. Side = √(8² + 6²) = 10 cm. Area = ½ × 16 × 12 = 96 cm². Perimeter 40 cm.

Problem 3 — a point inside a rectangle. O is any point inside rectangle ABCD. Prove OB² + OD² = OA² + OC². Draw PQ through O parallel to AB with P on AD and Q on BC. Then PQ ⊥ AD and PQ ⊥ BC, and AP = BQ, PD = QC (ABQP and PQCD are rectangles). OB² = OQ² + BQ², OD² = OP² + PD², OA² = OP² + AP², OC² = OQ² + QC². So OB² + OD² = OQ² + OP² + BQ² + PD² and OA² + OC² = OP² + OQ² + AP² + QC². Since AP = BQ and PD = QC, the two sums are equal.

Problem 4 — the same for a point outside or on a side. The identity OA² + OC² = OB² + OD² holds for any point O in the plane of the rectangle, by the same proof with signed segments; the examination asks only for the interior case.

Problem 5 — a parallelogram. In parallelogram ABCD, prove AC² + BD² = AB² + BC² + CD² + DA². Apply Apollonius' theorem to ΔABD with median AO (O the midpoint of BD): AB² + AD² = 2AO² + 2BO² = 2(AC/2)² + 2(BD/2)² = (AC² + BD²)/2. So AC² + BD² = 2(AB² + AD²) = AB² + BC² + CD² + DA² since opposite sides are equal. The rhombus identity is the special case with all sides equal.

Problem 6 — a square and a point on the diagonal. P is on diagonal AC of square ABCD. Prove PB = PD. ΔABP ≅ ΔADP (AB = AD, AP common, ∠BAP = ∠DAP = 45°), so PB = PD. And PB² = PA² + AB² − 2PA·AB·cos 45° is Intermediate material; at this level the congruence suffices.

Problem 7 — the midpoint quadrilateral. The midpoints of the sides of any quadrilateral form a parallelogram (proved via the midpoint theorem in two triangles). If the original is a rectangle, the midpoint figure is a rhombus (its sides are all half the equal diagonals); if the original is a rhombus, the midpoint figure is a rectangle (its sides are parallel to the perpendicular diagonals); if the original is a square, it is a square.

Problem 8 — a kite. In kite ABCD with AB = AD and CB = CD, the diagonal AC is the perpendicular bisector of BD, so with BD = 24 and AC = 25 split as AO = 9, OC = 16: AB = √(81 + 144) = 15 and CB = √(256 + 144) = 20.

Every one of these uses the same three moves: draw the diagonals or a parallel through the point, find the right triangles, write Pythagoras and combine.

The optional exercise's numerical problems all reduce to one or two right triangles; the work is in drawing the right figure.

Problem 1 — a wire between poles. Two poles of heights 6 m and 11 m stand on a plane ground, 12 m apart. Find the distance between their tops. Draw a horizontal from the top of the shorter pole to the taller; it meets the taller pole 6 m up, leaving 5 m above. The right triangle has legs 12 and 5, so the distance is √(144 + 25) = 13 m.

Problem 2 — a guy wire. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire is taut? √(24² − 18²) = √(576 − 324) = √252 = 6√7 ≈ 15.87 m.

Problem 3 — a ladder. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall. √(100 − 64) = 6 m. If the same ladder is turned to the other side of the street and reaches a window 6 m high, its foot is 8 m from that wall, and the street is 14 m wide if the foot stays in one place.

Problem 4 — aeroplanes. 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 the two after 1½ hours? Distances 1500 km and 1800 km at right angles: √(1500² + 1800²) = 300√(25 + 36) = 300√61 ≈ 2343 km.

Problem 5 — a man walking. A man goes 10 m due east and then 24 m due north. Find his distance from the starting point. √(100 + 576) = √676 = 26 m. (10, 24, 26 is twice the 5, 12, 13 triple.)

Problem 6 — a tree broken by wind. A tree breaks at a height of 5 m and its top touches the ground 12 m from its base. Find the original height. The broken part is the hypotenuse of a 5-12 right triangle, 13 m; original height 5 + 13 = 18 m.

Problem 7 — a hypotenuse from a difference. The hypotenuse of a right triangle is 6 m more than twice the shortest side, and the third side is 2 m less than the hypotenuse. Find the sides. Let the shortest be x: hypotenuse 2x + 6, third side 2x + 4. x² + (2x + 4)² = (2x + 6)²: x² + 4x² + 16x + 16 = 4x² + 24x + 36, x² − 8x − 20 = 0, (x − 10)(x + 2) = 0, x = 10. Sides 10, 24, 26.

Problem 8 — an equilateral triangle's altitude. A triangular park is equilateral with side 60 m; find the altitude and area. Altitude = (√3/2) × 60 = 30√3 ≈ 51.96 m; area = ½ × 60 × 30√3 = 900√3 ≈ 1558.8 m².

Problem 9 — a diagonal of a cuboid (extension). The space diagonal of a room 12 m × 9 m × 8 m is √(144 + 81 + 64) = √289 = 17 m, by Pythagoras applied twice: floor diagonal 15, then √(225 + 64).

Sketch, mark the right angle, label the known sides, decide whether the unknown is a leg or the hypotenuse, and compute; then state the answer with its unit and, where a triple appears, note it.

Some optional problems place the similar triangles in a circle or in a figure with an angle bisector, anticipating the next chapters.

Intersecting chords. Chords AB and CD of a circle meet at P inside the circle. Then PA × PB = PC × PD. Join AC and BD; ∠APC = ∠DPB (vertically opposite) and ∠ACP = ∠DBP (angles in the same segment on arc AD). So ΔAPC ~ ΔDPB and PA/PD = PC/PB. For PA = 4, PB = 6, PC = 3: PD = 8.

Chords meeting outside. If the chords meet at P outside the circle, the same product PA × PB = PC × PD holds, now with ΔPAC ~ ΔPDB (common angle P, and ∠PAC = ∠PDB because ABDC is cyclic, the exterior angle equalling the interior opposite). For PA = 3, AB = 5 (so PB = 8), PC = 4: PD = 6, CD = 2.

Tangent and secant. If PT is a tangent and PAB a secant, PT² = PA × PB, by ΔPTA ~ ΔPBT (common angle P, and ∠PTA = ∠PBT by the alternate segment theorem). This is proved in the tangents chapter; here it is a further instance of "product = similarity".

The angle bisector. In ΔABC, the bisector of ∠A meets BC at D. Then BD/DC = AB/AC. Draw CE ∥ DA meeting BA produced at E. Then ∠BAD = ∠AEC (corresponding) and ∠DAC = ∠ACE (alternate), and since ∠BAD = ∠DAC, ∠AEC = ∠ACE, so AE = AC. In ΔBCE, DA ∥ CE gives BD/DC = BA/AE = AB/AC. For AB = 6, AC = 4, BC = 5: BD = 3, DC = 2.

A triangle in a semicircle. The angle in a semicircle is a right angle, so a triangle with a diameter as one side is right-angled, and the altitude from the right angle to the diameter satisfies h² = (product of the two parts of the diameter). This is the geometric mean construction: to construct √(ab), lay a and b end to end, draw the semicircle on the sum, and erect the perpendicular at the join.

Two circles and a common tangent (extension). Two circles of radii 5 and 3 with centres 10 apart: the direct common tangent has length √(10² − 2²) = √96 = 4√6, by the right triangle whose legs are the tangent and the difference of radii.

Similar triangles from a light source. A lamp at height H casts the shadow of a stick of height h placed at distance d from the lamp's base; the shadow extends s beyond the stick with H/(d + s) = h/s, so s = hd/(H − h). For H = 4, h = 1, d = 6: s = 2.

The pattern to remember: any statement that one product of two lengths equals another product of two lengths is a proportion in disguise, and a proportion comes from a pair of similar triangles. Find the pair, prove the similarity with a criterion, and cross-multiply.

The four-mark proofs of this chapter are marked on structure as much as on content. A model layout, using the proof that BD² = AD × DC when BD is the altitude to the hypotenuse:

Given: ΔABC with ∠ABC = 90°, BD ⊥ AC, D on AC.
To prove: BD² = AD × DC.
Figure: [right triangle with altitude drawn, all points labelled].
Proof:
1. In ΔADB, ∠ADB = 90°, so ∠DAB + ∠ABD = 90°. [angle sum]
2. ∠ABC = 90°, so ∠ABD + ∠DBC = 90°. [given]
3. From 1 and 2, ∠DAB = ∠DBC. [complements of the same angle]
4. In ΔADB and ΔBDC: ∠ADB = ∠BDC = 90° and ∠DAB = ∠DBC. So ΔADB ~ ΔBDC. [AA criterion]
5. Hence AD/BD = DB/DC. [corresponding sides of similar triangles]
6. Therefore BD² = AD × DC. [cross-multiplying] ∎

The reasons that are accepted. For angles: given; vertically opposite; alternate or corresponding angles with named parallels; angle sum of a triangle; complements of the same angle; angles in the same segment; angle in a semicircle; base angles of an isosceles triangle; opposite angles of a parallelogram. For sides: given; midpoint; sides of a rhombus/square; diagonals of a rectangle equal; diagonals of a parallelogram bisect each other; corresponding sides of similar triangles; corresponding sides of congruent triangles. For similarity: AA, SSS, SAS, with the correspondence stated. For lengths: Pythagoras in a named right triangle; the BPT with the parallel named; the area theorem.

Naming the correspondence. ΔADB ~ ΔBDC says A ↔ B, D ↔ D, B ↔ C. The proportion is then read off in order: AD/BD = DB/DC = AB/BC. Writing the triangles as ΔABD ~ ΔBDC would give a false proportion. When unsure, list the three pairs of equal angles and match the vertices by their angles.

Constructions in proofs. The converse of the BPT needs the line DE' ∥ BC; the converse of Pythagoras needs the triangle PQR with a right angle; the trapezium proof by BPT needs EF through O; the area theorem needs the two altitudes; Apollonius needs the altitude from A; the rectangle problem needs the parallel through O. State the construction before the proof and mark it on the figure.

Checking a proof. Every line must follow from earlier lines or from a stated fact. Circular reasoning — using what is to be proved — is the most common invisible error; the second is using a similarity that has not been established. Read the proof backwards from the conclusion: each step should have a reason to its right.

Time and marks. A four-mark proof should take six to eight minutes. Figure and setup: one mark. The key similarity or Pythagoras statements with reasons: two marks. The algebra and conclusion: one mark. A figure that contradicts the statement (a right angle drawn at the wrong vertex) can lose all four.

Practise the six named proofs of the chapter — BPT, its converse, the area theorem, Pythagoras, its converse, and the trapezium diagonals — until each can be written from memory with the figure. The optional problems are variations on them.

A set of problems in the exercise's own style, each solved fully.

Problem 1. In ΔABC, D is the midpoint of BC and E is the midpoint of AD. BE produced meets AC at F. Prove AF = ⅓ AC. Draw DG ∥ BF meeting AC at G. In ΔADG, E is the midpoint of AD and EF ∥ DG, so F is the midpoint of AG: AF = FG. In ΔBCF, D is the midpoint of BC and DG ∥ BF, so G is the midpoint of FC: FG = GC. Hence AF = FG = GC and AF = ⅓ AC.

Problem 2. ABCD is a parallelogram and P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given area(ΔCPQ) = 20 cm², find area(ΔCPD) and area(ΔBPQ). ΔBPQ ~ ΔCPD (bow-tie: vertically opposite at P, alternate angles from AB ∥ DC), with BP/CP = 1/2, so area(BPQ)/area(CPD) = 1/4. Also ΔCPQ and ΔCPD share the vertex C and have bases PQ and PD on one line, and PQ/PD = BP/CP = 1/2, so area(CPD) = 2 × area(CPQ) = 40 cm². Hence area(BPQ) = 10 cm².

Problem 3. In ΔABC, AB = AC and D is a point on BC. Prove AB² − AD² = BD × DC. Draw AE ⊥ BC; E is the midpoint of BC since the triangle is isosceles. AB² = AE² + BE² and AD² = AE² + DE². Subtract: AB² − AD² = BE² − DE² = (BE − DE)(BE + DE) = BD × (CE + DE) = BD × DC, using BE = CE.

Problem 4. In ΔABC, ∠A = 90° and AD ⊥ BC. If BD = 8 and DC = 2, find AD, AB and AC. AD² = BD × DC = 16, AD = 4. AB² = BD × BC = 8 × 10 = 80, AB = 4√5. AC² = DC × BC = 20, AC = 2√5. Check: 80 + 20 = 100 = BC².

Problem 5. Two poles of heights a and b stand on level ground at distance p apart. A wire from the top of each to the foot of the other; the wires cross at height h. Prove 1/h = 1/a + 1/b. Let the crossing be at horizontal distance x from the pole of height a. Similar triangles give h/b = x/p and h/a = (p − x)/p. Adding: h/a + h/b = 1, so 1/h = 1/a + 1/b. For a = 6, b = 3: h = 2, independent of p.

Problem 6. ABC is an isosceles triangle right-angled at C. Prove AB² = 2AC². By Pythagoras AB² = AC² + BC², and BC = AC, so AB² = 2AC².

Problem 7. In ΔPQR, PD ⊥ QR with D on QR, and PQ = a, PR = b, QD = c, DR = d. Prove (a + b)(a − b) = (c + d)(c − d). a² = PD² + c² and b² = PD² + d²; subtract: a² − b² = c² − d², which factorises to the required.

Problem 8. The sides of a triangle are 2x, 2x + 2 and 2x + 4 with x > 0, and it is right-angled. Find x. (2x)² + (2x + 2)² = (2x + 4)²: 4x² + 4x² + 8x + 4 = 4x² + 16x + 16, 4x² − 8x − 12 = 0, x² − 2x − 3 = 0, (x − 3)(x + 1) = 0, x = 3. Sides 6, 8, 10.

Each problem was solved by one of the chapter's standard moves; the skill practised is choosing the move quickly.

Decide the type. A product of two segments equals another product → find similar triangles and cross-multiply. A relation among squares of sides → find right triangles and write Pythagoras. A ratio of areas → find similar triangles and square the side ratio, or use same-base-same-height. A ratio of segments on a side → BPT with a named parallel. A claim that a line is parallel → converse BPT. A claim of a right angle → converse Pythagoras.

Find the pair. Look for the three shapes: nested (a parallel to a side), bow-tie (parallel sides across a crossing), altitude-to-hypotenuse. If none is visible, construct one: a parallel through a point, an altitude, a line through a midpoint. Name the two triangles with vertices in corresponding order.

Prove and compute. State the criterion with the equal angles or proportional sides and their reasons. Write the proportion. Cross-multiply or substitute. For Pythagoras, write the equation for each right triangle, then add or subtract to remove the segment not wanted.

Check. Test the identity with a specific triangle: 3-4-5 for right triangles, side 6 for equilateral, AB = 2CD for the trapezium. If the numbers fail, the proof has an error, usually in the correspondence or in a sign.

Common errors. Wrong correspondence in a similarity. Using AA with angles that are not corresponding. Applying the BPT to a non-parallel line. Adding areas of triangles that overlap. Forgetting that the area ratio is the square of the side ratio. Mixing units. Assuming the figure's appearance (a right angle that is not given). Using the result to be proved as a step.

Where this leads. The next two chapters — Tangents and Secants to a Circle, and Trigonometry — depend entirely on similar right triangles. The tangent–secant theorem PT² = PA × PB is a similarity; the trigonometric ratios are defined as ratios of sides of a right triangle and are well-defined only because similar right triangles have equal ratios. The Intermediate course uses similarity in the geometry of the circle, in vectors, and in the calculus of related rates; the habit of reducing a figure to right triangles serves in physics from projectile motion to optics.

The optional exercise is not an extra; it is the chapter's content used the way it will be used later. A student who can write the six named proofs and solve the product, square and area families without hesitation is prepared for every geometry question of the SSC paper and for what follows.