Chapter 6 — Triangles

In Class 9 you learnt that two geometrical figures are congruent if they have exactly the same shape and the same size, so that one can be placed over the other to cover it completely. All circles of radius 3 cm are congruent; two squares of side 5 cm are congruent. But look at two circles of radii 3 cm and 5 cm, or two squares of sides 2 cm and 7 cm. They are not congruent, since their sizes differ, yet their shapes are exactly the same. Such figures are called similar figures.

Two figures are similar if they have the same shape but not necessarily the same size. Observe carefully that every pair of congruent figures is also a pair of similar figures, but similar figures need not be congruent. Congruence is a special case of similarity in which the size also matches.

Some facts follow immediately. All circles are similar to one another. All squares are similar. All equilateral triangles are similar. In fact, any two regular polygons with the same number of sides are similar. But two rectangles need not be similar: a rectangle of 2 cm by 10 cm is long and thin, while a rectangle of 4 cm by 5 cm is nearly square, and no enlargement of one produces the other. Two quadrilaterals are similar only when a definite condition is satisfied.

For two polygons with the same number of sides, similarity requires both of the following at the same time:

• their corresponding angles are equal, and
• their corresponding sides are in the same ratio (that is, proportional).

The common ratio of corresponding sides is called the scale factor or the representative fraction. A map drawn on a scale of 1 : 100000 is similar to the region it represents with scale factor 1/100000. When a photograph is enlarged from 4 cm by 6 cm to 8 cm by 12 cm, every length is doubled and every angle stays the same, so the two prints are similar with scale factor 2.

Why do we need both conditions? Consider a square and a rectangle that is not a square. All their angles are 90°, so corresponding angles are equal, but the sides are not proportional; the figures are not similar. Now consider a square and a rhombus that is not a square. All four sides of each are equal, so the sides are proportional, but the angles differ; again the figures are not similar. Either condition alone is not enough for polygons in general. Remarkably, for triangles, as we shall see, one condition forces the other, and that is what makes the study of similar triangles so neat.

A triangle is the simplest polygon, and the definition of similar polygons applies to it directly. Two triangles are similar if (i) their corresponding angles are equal, and (ii) their corresponding sides are in the same ratio. The Greek mathematician Thales is credited with first observing that the ratio of any two corresponding sides in two equiangular triangles is always the same.

We write similarity with the symbol ~. If triangle ABC is similar to triangle DEF we write ΔABC ~ ΔDEF, and this means all of the following at once:

• ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, and
• AB/DE = BC/EF = CA/FD.

The order of letters matters. The statement ΔABC ~ ΔDEF pairs A with D, B with E and C with F. The same two triangles written as ΔABC ~ ΔEFD would pair A with E, which may be a false statement. When you solve a problem, first identify which angle of one triangle equals which angle of the other, then write the vertices in that matching order, and only then write the ratio of sides by reading corresponding letters in the two names. This single habit removes most of the errors students make in this chapter.

Corresponding sides lie opposite corresponding angles. In ΔABC ~ ΔDEF, side BC is opposite ∠A and side EF is opposite ∠D, so BC and EF correspond. The common value of the ratios of corresponding sides is the scale factor; if it is 1 the triangles are congruent.

A very important fact, which the rest of the chapter proves in stages, is that for triangles the two conditions in the definition are not independent. If the three angles of one triangle equal the three angles of another, the sides are automatically proportional; and if the three sides are proportional, the angles are automatically equal. This is not true for quadrilaterals or other polygons, as the square and rhombus example showed. It is true for triangles because a triangle is rigid: three sides fix its shape completely, and two angles fix its shape up to size.

Since the angles of a triangle add up to 180°, if two angles of one triangle equal two angles of another, the third angles are also equal. This observation will give us the AA criterion, the most frequently used test of similarity in examination problems.

The whole theory of similar triangles rests on one theorem about a line drawn parallel to one side of a triangle. It is called the Basic Proportionality Theorem, or Thales theorem.

Theorem. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Given: ΔABC in which 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 DM ⊥ AC and EN ⊥ AB.

Proof. Area of a triangle is ½ × base × height. Taking AD as the base of ΔADE with height EN,
ar(ADE) = ½ × AD × EN, and ar(BDE) = ½ × DB × EN.
Therefore ar(ADE)/ar(BDE) = AD/DB. … (1)

Now take AE as the base of ΔADE with height DM:
ar(ADE) = ½ × AE × DM, and ar(DEC) = ½ × EC × DM.
Therefore ar(ADE)/ar(DEC) = AE/EC. … (2)

Triangles BDE and DEC stand on the same base DE and lie between the same parallels DE and BC. Hence ar(BDE) = ar(DEC). … (3)

From (1), (2) and (3), AD/DB = AE/EC. This proves the theorem.

The theorem can be used in several equivalent forms, all obtained by simple algebra. If AD/DB = AE/EC, then adding 1 to both sides gives (AD + DB)/DB = (AE + EC)/EC, that is, AB/DB = AC/EC. Taking reciprocals and rearranging also gives AD/AB = AE/AC and DB/AB = EC/AC. In a problem choose whichever form contains the three known lengths and the one unknown.

Notice the role of the construction: the two extra perpendiculars let us express each area with two different bases and heights, and the fact that BDE and DEC have equal areas because of the parallel lines is the link between the two ratios. In the board examination the proof is asked as a four- or five-mark question and full marks require the figure, the given, the to-prove, the construction and each step with its reason.

A theorem says that a condition leads to a result; its converse asks whether the result leads back to the condition. For the Basic Proportionality Theorem the converse is also true and is equally useful.

Theorem (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.

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

Proof. Suppose DE is not parallel to BC. Then through D draw a line DE′ parallel to BC, meeting AC at E′ (E′ different from E). By the Basic Proportionality Theorem applied to DE′ ∥ BC,
AD/DB = AE′/E′C. … (1)
But we are given AD/DB = AE/EC. … (2)
From (1) and (2), AE′/E′C = AE/EC. Adding 1 to both sides, (AE′ + E′C)/E′C = (AE + EC)/EC, that is, AC/E′C = AC/EC. Hence E′C = EC, so E′ and E coincide. This contradicts our supposition that E′ is different from E. Therefore DE must be parallel to BC.

This is a proof by contradiction: we assume the opposite of what we want, use a known theorem, and arrive at an impossibility. The board expects you to state clearly where the contradiction arises.

The converse is the tool for proving that two lines are parallel. Typical uses are:

• Showing that the line joining the mid-points of two sides of a triangle is parallel to the third side (mid-point theorem): if D and E are mid-points then AD/DB = 1 = AE/EC, so DE ∥ BC.
• Proving that a quadrilateral is a trapezium by showing one pair of sides parallel.
• Checking whether a given DE is parallel to BC from measured lengths.

A frequently asked result combines the theorem and its converse. In trapezium ABCD with AB ∥ DC, a line through the intersection O of the diagonals, or any line parallel to AB, meets AD at E and BC at F; then AE/ED = BF/FC. The proof joins A to C, meeting EF at G, and applies BPT twice, in ΔADC to EG ∥ DC and in ΔCAB to GF ∥ AB. Learn this argument; it is a standard three-mark question.

Examination questions on the Basic Proportionality Theorem are of three kinds: find an unknown length, check whether a line is parallel, and prove a ratio result in a figure that contains one or more parallel lines. Let us work through the method of each.

Finding a length. Identify the triangle, the side to which the given line is parallel, and the two sides that are cut. Write the proportion in the form that contains the unknown once. Suppose in ΔABC, DE ∥ BC, AD = x, DB = x − 2, AE = x + 2 and EC = x − 1. By BPT, AD/DB = AE/EC, so x/(x − 2) = (x + 2)/(x − 1). Cross-multiplying, x(x − 1) = (x + 2)(x − 2), that is, x² − x = x² − 4, giving x = 4. Then AD = 4, DB = 2, AE = 6, EC = 3, and indeed 4/2 = 6/3.

Checking parallelism. Compute the two ratios AD/DB and AE/EC as fractions in lowest terms. If they are equal, the line is parallel by the converse; if not, it is not parallel. Be careful to use the segments into which the point divides the side, not the whole side, unless you deliberately use the form AD/AB = AE/AC.

Proving a result. When the figure contains two triangles that share a vertex and each has a line parallel to its base, apply BPT in each triangle and compare. A classic is the following. In ΔABC, D is a point on AB, E and F are points on BC with DE ∥ AC and DF ∥ AE; prove that BF/FE = BE/EC. In ΔBAE, DF ∥ AE gives BD/DA = BF/FE; in ΔBAC, DE ∥ AC gives BD/DA = BE/EC; hence BF/FE = BE/EC.

Another standard problem uses the converse. Given a quadrilateral ABCD whose diagonals intersect at O with AO/BO = CO/DO, prove ABCD is a trapezium: draw OE ∥ AB through O meeting AD at E; in ΔDAB, EO ∥ AB gives DE/EA = DO/OB; the given condition gives DO/OB = CO/AO; so DE/EA = CO/AO, and by the converse of BPT in ΔADC, EO ∥ DC. Since EO ∥ AB, we get AB ∥ DC.

In every case write the theorem you are using, the triangle in which you are using it, and the parallel line, as a bracketed reason next to the step.

To use the definition of similar triangles we would have to check three pairs of angles and three ratios of sides. The criteria of similarity reduce this work; they are the similarity counterparts of the congruence rules SSS, SAS, ASA of Class 9.

Theorem (AAA criterion). If in two triangles the corresponding angles are equal, then their corresponding sides are in the same ratio, and hence the two triangles are similar.

Given: ΔABC and ΔDEF with ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
To prove: AB/DE = BC/EF = AC/DF.
Construction: Suppose DE is longer than AB. Cut DP = AB on DE and DQ = AC on DF and join PQ.

Proof. In ΔABC and ΔDPQ, AB = DP, AC = DQ and ∠A = ∠D (given). So ΔABC ≅ ΔDPQ by SAS congruence. Therefore ∠B = ∠DPQ. But ∠B = ∠E (given), so ∠DPQ = ∠E. These are corresponding angles for the lines PQ and EF with transversal DE, hence PQ ∥ EF. By the Basic Proportionality Theorem in ΔDEF, DP/PE = DQ/QF, which gives DP/DE = DQ/DF, that is, AB/DE = AC/DF. Similarly, by cutting off segments on ED and EF, AB/DE = BC/EF. Hence AB/DE = BC/EF = AC/DF and ΔABC ~ ΔDEF.

Since the sum of the angles of a triangle is 180°, if two angles of one triangle are equal to two angles of another, the third pair is automatically equal. So the AAA criterion is in practice the AA criterion: two triangles are similar if two angles of one are respectively equal to two angles of the other.

The AA criterion is the one used most often because equal angles are easy to find in figures: vertically opposite angles, common angles shared by two overlapping triangles, right angles, alternate angles between parallel lines, angles in the same segment of a circle, and angles of equilateral triangles. When two triangles overlap and share an angle, look for one more pair of equal angles and the similarity follows.

Once similarity is established, write the correspondence carefully. If ∠A = ∠D and ∠B = ∠E, then ΔABC ~ ΔDEF and the ratios are AB/DE, BC/EF, CA/FD. Writing ΔABC ~ ΔEDF would silently exchange sides and give wrong lengths.

The second criterion goes in the opposite direction: from sides to angles.

Theorem (SSS criterion). If in two triangles the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Given: ΔABC and ΔDEF with AB/DE = BC/EF = CA/FD (each ratio less than 1, say).
To prove: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, so ΔABC ~ ΔDEF.
Construction: Cut DP = AB on DE and DQ = AC on DF; join PQ.

Proof. Since AB/DE = AC/DF, we have DP/DE = DQ/DF, which gives DP/PE = DQ/QF. By the converse of BPT, PQ ∥ EF. Therefore ∠P = ∠E and ∠Q = ∠F (corresponding angles). So ΔDPQ ~ ΔDEF by AA, giving DP/DE = PQ/EF. But DP/DE = AB/DE = BC/EF (given), so PQ/EF = BC/EF, hence PQ = BC. Now in ΔABC and ΔDPQ, AB = DP, BC = PQ and CA = QD, so ΔABC ≅ ΔDPQ by SSS congruence. Therefore ∠A = ∠D, ∠B = ∠P = ∠E and ∠C = ∠Q = ∠F. Hence ΔABC ~ ΔDEF.

The pattern of the proof, cutting off a triangle congruent to the smaller one inside the larger one and using BPT with its converse, is the same as in the AAA proof. Understanding this pattern lets you reproduce either proof without memorising.

To apply the SSS criterion to numbers, arrange the sides of each triangle in increasing order, pair the smallest with the smallest, the largest with the largest, and check that the three ratios are equal. The equal angles then lie opposite the paired sides: the largest angle of each triangle is opposite its largest side. This is how you decide the correspondence in an SSS question, because no angles are given.

A very common mistake is to test only two ratios. All three must be equal. Sides 3, 4, 6 and 6, 8, 9 give 3/6 = 4/8 = 1/2 but 6/9 = 2/3, so the triangles are not similar.

The third criterion uses two sides and the angle between them, exactly as SAS congruence does, but with proportional sides instead of equal sides.

Theorem (SAS criterion). If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

Given: ΔABC and ΔDEF with ∠A = ∠D and AB/DE = AC/DF.
To prove: ΔABC ~ ΔDEF.
Construction: Cut DP = AB on DE and DQ = AC on DF; join PQ.

Proof. In ΔABC and ΔDPQ, AB = DP, AC = DQ and ∠A = ∠D, so ΔABC ≅ ΔDPQ by SAS congruence; hence ∠B = ∠DPQ and ∠C = ∠DQP. Also AB/DE = AC/DF gives DP/DE = DQ/DF, so DP/PE = DQ/QF and by the converse of BPT, PQ ∥ EF. Therefore ∠DPQ = ∠E and ∠DQP = ∠F (corresponding angles). Combining, ∠B = ∠E and ∠C = ∠F, and with ∠A = ∠D the triangles ΔABC and ΔDEF are equiangular, hence similar by AAA.

The word including is essential. The angle must be the one formed by the two sides whose ratio is taken. If ∠A = ∠D but the proportional sides are AB/DE and BC/EF, the angle A is not included between AB and BC in the sense required, because BC does not pass through A; the criterion does not apply and the triangles may not be similar. This is the same reason there is no SSA rule for congruence.

SAS is the criterion to use when a point divides two sides of a triangle in the same ratio and the triangles share the angle at the common vertex. If D on AB and E on AC satisfy AD/AB = AE/AC, then ΔADE ~ ΔABC by SAS (common ∠A), which gives another way of seeing that DE ∥ BC and DE/BC = AD/AB.

Summary of the three criteria: AA (two angles), SSS (three sides proportional), SAS (two sides proportional and the included angle equal). Together with the Basic Proportionality Theorem they solve every problem in this chapter. In a proof question, name the criterion you use and list the three facts it needs.

Once the three criteria are known, most chapter questions follow a fixed routine: identify two triangles, prove them similar by naming a criterion, write the correct correspondence, and then extract the ratio of sides needed. Let us practise the routine on the standard examination problems.

Problem 1. In the figure, ∠ACB = 90° and CD ⊥ AB. Prove CD² = BD × AD. In ΔACD and ΔCBD, ∠ADC = ∠CDB = 90°. Since ∠ACD + ∠DCB = 90° and ∠DCB + ∠B = 90° (angles of the right triangle CDB), we get ∠ACD = ∠B. So ΔACD ~ ΔCBD by AA. Then CD/BD = AD/CD, giving CD² = AD × BD.

Problem 2. ABC and AMP are two right triangles, right-angled at B and M respectively, with M on AC and P on the line through A. Prove ΔABC ~ ΔAMP and CA/PA = BC/MP. In ΔABC and ΔAMP, ∠ABC = ∠AMP = 90° and ∠A is common, so ΔABC ~ ΔAMP by AA; reading the corresponding sides, CA/PA = BC/MP.

Problem 3. E is a point on side AD produced of parallelogram ABCD and BE meets CD at F. Show ΔABE ~ ΔCFB. ∠A = ∠C (opposite angles of a parallelogram) and ∠AEB = ∠CBF (alternate angles, AD ∥ BC with transversal BE). Hence ΔABE ~ ΔCFB by AA.

Problem 4. Sides AB and BC and median AD of ΔABC are proportional to sides PQ and QR and median PM of ΔPQR. Show ΔABC ~ ΔPQR. Given AB/PQ = BC/QR = AD/PM. Since D and M are mid-points, BC/QR = BD/QM, so AB/PQ = BD/QM = AD/PM and ΔABD ~ ΔPQM by SSS, giving ∠B = ∠Q. Now AB/PQ = BC/QR with the included angles ∠B = ∠Q, so ΔABC ~ ΔPQR by SAS.

Problem 5. Two poles of heights 6 m and 11 m stand on level ground 12 m apart; find the distance between their tops. Draw the horizontal from the top of the shorter pole; it forms a right triangle with legs 12 m and 11 − 6 = 5 m, so the distance is √(144 + 25) = 13 m.

The habit to build: before writing any ratio, write the similarity statement with vertices in matching order and then copy the ratios from it letter by letter. In the ratio AB/DE = BC/EF = CA/FD the first letters A, B, C of one name pair with D, E, F of the other in the same positions.

A right triangle has a special structure that similarity reveals beautifully. Draw the perpendicular from the right-angle vertex to the hypotenuse; it splits the triangle into two smaller right triangles, and each of them is similar to the whole triangle and to the other.

Theorem. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Given: ΔABC right-angled at B, and BD ⊥ AC with D on AC.
To prove: ΔADB ~ ΔABC, ΔBDC ~ ΔABC and ΔADB ~ ΔBDC.

Proof. In ΔADB and ΔABC, ∠ADB = ∠ABC = 90° and ∠A is common; so ΔADB ~ ΔABC by AA. In ΔBDC and ΔABC, ∠BDC = ∠ABC = 90° and ∠C is common; so ΔBDC ~ ΔABC by AA. Since both small triangles are similar to ΔABC, they are similar to each other: ΔADB ~ ΔBDC.

Write the correspondences precisely. From ΔADB ~ ΔABC: AD/AB = DB/BC = AB/AC, so AB² = AD × AC. From ΔBDC ~ ΔABC: BD/AB = DC/BC = BC/AC, so BC² = CD × AC. From ΔADB ~ ΔBDC: AD/BD = DB/DC, so BD² = AD × DC. These three results are examined again and again. In words: each leg squared equals the product of the hypotenuse and the projection of that leg on the hypotenuse, and the altitude squared equals the product of the two segments of the hypotenuse.

Adding the first two results gives AB² + BC² = AD × AC + CD × AC = (AD + CD) × AC = AC × AC = AC². This is exactly the Pythagoras theorem, and it is how the chapter proves it in the next topics.

A useful numerical consequence: if the legs are a, b and the hypotenuse is c, then the altitude to the hypotenuse is h = ab/c, because the area of the triangle can be computed two ways, ½ab = ½ch. For a 3-4-5 triangle, h = 12/5 = 2.4, and the hypotenuse segments are 9/5 and 16/5, whose product 144/25 equals h², as the theorem predicts.

Similar triangles have their sides in a fixed ratio; what is the ratio of their areas? Since area involves the product of two lengths, it is natural to expect the square of the side ratio, and that is exactly the theorem.

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: ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)².
Construction: Draw altitudes AM ⊥ BC and PN ⊥ QR.

Proof. ar(ABC) = ½ × BC × AM and ar(PQR) = ½ × QR × PN, so ar(ABC)/ar(PQR) = (BC × AM)/(QR × PN). … (1)
In ΔABM and ΔPQN, ∠B = ∠Q (since ΔABC ~ ΔPQR) and ∠M = ∠N = 90°, so ΔABM ~ ΔPQN by AA; hence AM/PN = AB/PQ. … (2)
Also, from the similarity of the given triangles, AB/PQ = BC/QR. … (3)
From (1), (2) and (3): ar(ABC)/ar(PQR) = (BC/QR) × (AM/PN) = (BC/QR) × (BC/QR) = (BC/QR)². Using (3) again the same ratio equals (AB/PQ)² and (CA/RP)².

The theorem applies to any pair of corresponding linear measurements, not just sides: the ratio of areas also equals the square of the ratio of corresponding altitudes, medians, angle bisectors or perimeters, since each of these is in the same ratio as the sides.

Two consequences are tested frequently. First, if two similar triangles have equal areas, the ratio of sides is 1, so the triangles are congruent. Second, if D, E, F are the mid-points of the sides of ΔABC, then ΔDEF ~ ΔABC with ratio ½ (mid-point theorem), so ar(DEF) : ar(ABC) = 1 : 4.

In numerical work, take square roots carefully when going from areas to sides: if areas are in the ratio 81 : 49, the sides are in the ratio 9 : 7, not 81 : 49. And when going from sides to areas, square the ratio: sides 4 : 9 give areas 16 : 81.

You verified the Pythagoras theorem in earlier classes by counting squares or by cutting paper. Now, using similarity of right triangles, we can prove it rigorously. In Indian mathematics the statement appears in the Baudhayana Sulba Sutra, centuries before Pythagoras, so it is also called the Baudhayana theorem.

Theorem. 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.

Proof. ΔADB ~ ΔABC (the two triangles on either side of the altitude are similar to the whole triangle), so AD/AB = AB/AC, that is, AD × AC = AB². … (1)
ΔBDC ~ ΔABC, so CD/BC = BC/AC, that is, CD × AC = BC². … (2)
Adding (1) and (2): AD × AC + CD × AC = AB² + BC², so (AD + CD) × AC = AB² + BC². Since AD + CD = AC, we get AC × AC = AB² + BC², that is, AC² = AB² + BC².

The theorem lets us find any side of a right triangle from the other two, and it identifies many important lengths: the diagonal of a rectangle, the height of an equilateral triangle, the distance between two points. Certain sets of whole numbers, called Pythagorean triplets, satisfy the relation: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (9, 40, 41), and any multiple of these such as (6, 8, 10) or (10, 24, 26). Recognising a triplet saves time in the examination.

Two results used constantly follow directly. In an equilateral triangle of side a, the altitude bisects the base, so h² = a² − (a/2)² = 3a²/4 and h = (√3/2)a. In a square of side a, the diagonal is a√2. And the altitude of an equilateral triangle gives the frequently examined statement that three times the square of a side equals four times the square of the altitude: 3a² = 4h².

A more general fact worth knowing: in ΔABC with AD ⊥ BC, we always have AB² − BD² = AC² − CD², since both sides equal AD².

The Pythagoras theorem tells us what is true in a right triangle. Its converse lets us decide whether a triangle is right-angled just from the lengths of its sides, which is exactly what a carpenter or a surveyor needs when setting out a right angle with a rope.

Theorem (converse of Pythagoras). 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: Construct ΔPQR right-angled at Q such that PQ = AB and QR = BC.

Proof. In ΔPQR, by the Pythagoras theorem, PR² = PQ² + QR² = AB² + BC² (by construction). … (1)
But AC² = AB² + BC² (given). … (2)
From (1) and (2), PR² = AC², so PR = AC. Now in ΔABC and ΔPQR, AB = PQ, BC = QR and AC = PR, so ΔABC ≅ ΔPQR by SSS congruence. Hence ∠B = ∠Q = 90°.

The procedure for a numerical question is fixed: take the largest side, square it, and compare with the sum of the squares of the other two sides. If they are equal, the triangle is right-angled and the right angle is opposite the largest side. If the square of the largest side is greater, the largest angle is obtuse; if smaller, all angles are acute. Only the equality is part of the syllabus statement, but the comparison helps you reason.

Ancient Indian builders used a rope with knots dividing it in the ratio 3 : 4 : 5 to lay out right angles for altars, and a modern mason still uses the 3-4-5 rule to check the corner of a wall. A rope divided into 12 equal parts, pulled taut into a triangle with sides of 3, 4 and 5 parts, gives an exact right angle by this converse.

A standard proof-question that combines both theorems: in an isosceles triangle ABC with AC = BC, if AB² = 2AC², prove ∠C = 90°. Since AB² = AC² + AC² = AC² + BC², the converse gives ∠C = 90° directly.

The practical value of similar triangles is that a length we cannot measure directly can be found from lengths we can measure. The Sun's rays are parallel, so at any instant a vertical pole and its shadow, and a tower and its shadow, form two similar right triangles. Thales is said to have measured the height of a pyramid in Egypt by this method, comparing the pyramid's shadow with the shadow of a stick.

Shadow method. A vertical stick of height h casts a shadow of length s at the same time as a tower of unknown height H casts a shadow of length S. The two right triangles have the same angle of elevation of the Sun, so they are similar by AA, and H/h = S/s, giving H = hS/s. For a 2 m stick with a 3 m shadow, and a tower shadow of 45 m, H = 2 × 45/3 = 30 m.

Mirror method. A mirror is placed flat on the ground between an observer and a tree. The observer moves until the top of the tree is seen in the mirror. Since the angle of incidence equals the angle of reflection, the angle at the mirror is the same on both sides, and both triangles have a right angle at the feet, so they are similar. If the observer's eye is 1.5 m high and stands 2 m from the mirror, and the mirror is 20 m from the foot of the tree, the tree's height is 1.5 × 20/2 = 15 m.

Lamp post and shadow. A person of height p walks away from a lamp of height L. When the person is at distance d from the post, the shadow of length x satisfies L/p = (d + x)/x, by the similarity of the big triangle (lamp, tip of shadow) and the small triangle (person, tip of shadow). Solving, x = pd/(L − p). Note that the length of the shadow is proportional to d: the tip of the shadow moves at a constant multiple of the person's speed, namely L/(L − p) times.

Width of a river. Stand at a point A on one bank opposite a tree T on the other bank. Walk a known distance AB along the bank, plant a stick at B, walk a further distance BC, and then walk away from the bank at right angles to a point D from which the stick B and the tree T are in a straight line. Then ΔTAB ~ ΔDCB (right angles at A and C, vertically opposite angles at B), and TA/DC = AB/CB, giving the width TA = DC × AB/CB.

In every such problem, draw the figure, mark the right angles and the equal angles, state the similarity with vertices in matching order, and only then write the proportion.