Thales’ theorem states that an angle inscribed in a semicircle is a right angle: if BC is a diameter of a circle and A is any other point on the circle, then ∠BAC = 90°. That is the meaning used internationally. In Indian school maths, though, the NCERT Class 10 chapter on Triangles uses “Thales theorem” for a different result, the Basic Proportionality Theorem (BPT): a line drawn parallel to one side of a triangle divides the other two sides in the same ratio. Both are credited to Thales of Miletus (about 624 to 546 BCE), so this page covers both, with proofs, converses and solved examples.
| Thales’ theorem (semicircle) | Basic Proportionality Theorem (NCERT “Thales theorem”) | |
|---|---|---|
| Statement | Angle in a semicircle = 90° | DE ∥ BC gives AD/DB = AE/EC |
| Topic | Circles | Similar triangles |
| Where taught in India | Circle geometry (“angle in a semicircle”) | Class 10 Triangles, Theorem 6.1 |
| Standard proof | Two isosceles triangles | Ratio of triangle areas |
| Converse | A right angle subtends a diameter | Equal ratios mean the line is parallel |
Thales’ theorem: the angle in a semicircle
Statement: If A, B and C are distinct points on a circle and BC is a diameter, then ∠BAC = 90°.
Thales theorem proof (isosceles triangles)
- Let O be the centre of the circle, so O is the midpoint of the diameter BC. Join OA.
- OA, OB and OC are all radii, so OA = OB = OC.
- In triangle OAB, OA = OB, so it is isosceles and its base angles are equal: ∠OBA = ∠OAB. Call this angle α.
- In triangle OAC, OA = OC, so ∠OCA = ∠OAC. Call this angle β.
- In the big triangle ABC, the angles are ∠B = α, ∠C = β and ∠A = ∠OAB + ∠OAC = α + β.
- Angle sum of triangle ABC: α + β + (α + β) = 180°, so 2(α + β) = 180° and α + β = 90°.
- Therefore ∠BAC = α + β = 90°. Proved.
It is also a special case of the inscribed angle theorem: the angle at the centre is twice the angle at the circumference, and a diameter makes a straight angle of 180° at the centre, so the inscribed angle is 90°.
Converse of Thales’ theorem
If ∠BAC = 90°, then A lies on the circle with BC as diameter. Equivalently, the circumcentre of a right triangle is the midpoint of its hypotenuse, and the circumradius is half the hypotenuse. Proof sketch: complete the rectangle ABDC by reflecting A through the midpoint M of BC. The diagonals of a rectangle are equal and bisect each other, so MA = MB = MC, and A is on the circle of centre M and radius BC/2.
Worked example: semicircle
BC is a diameter of a circle, BC = 13 cm, and A is a point on the circle with AB = 5 cm. Find AC and the radius.
- By Thales’ theorem, ∠BAC = 90°, so BC is the hypotenuse of right triangle ABC.
- By the Pythagorean theorem, AC² = BC² − AB² = 169 − 25 = 144, so AC = 12 cm.
- Radius = BC/2 = 6.5 cm. Check: the midpoint of the hypotenuse is 6.5 cm from A, B and C, as the converse predicts.
Basic Proportionality Theorem (Thales theorem in NCERT)
Statement (NCERT Class 10, Theorem 6.1): 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. In triangle ABC with D on AB, E on AC and DE ∥ BC:
AD/DB = AE/EC
Outside India the same idea is usually called the intercept theorem or side-splitter theorem.
BPT proof using areas (the NCERT method)
Given: triangle ABC, DE ∥ BC, D on AB, E on AC. To prove: AD/DB = AE/EC.
Construction: join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
- Area of a triangle = ½ × base × height. Take AD and DB as bases with the common height EN:
ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN.
So ar(ADE)/ar(BDE) = AD/DB. (i) - Now take AE and EC as bases with the common height DM:
ar(ADE) = ½ × AE × DM and ar(DEC) = ½ × EC × DM.
So ar(ADE)/ar(DEC) = AE/EC. (ii) - Triangles BDE and DEC stand on the same base DE and lie between the same parallels DE and BC, so ar(BDE) = ar(DEC). (iii)
- From (i), (ii) and (iii), the left-hand sides of (i) and (ii) are equal, so AD/DB = AE/EC. Proved.
Two useful consequences follow by adding 1 to both sides or inverting: AB/AD = AC/AE and AB/DB = AC/EC. And since triangle ADE is similar to triangle ABC, DE/BC = AD/AB, which the proportion alone does not give you (DE/BC is not equal to AD/DB).
Converse of the Basic Proportionality Theorem
NCERT Theorem 6.2: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side. The standard proof is by contradiction: if DE were not parallel to BC, draw DE′ ∥ BC through D. By BPT, AD/DB = AE′/E′C, and combined with the given AD/DB = AE/EC this forces E and E′ to divide AC in the same ratio, so they coincide. Hence DE ∥ BC.
Worked example: BPT
In triangle ABC, DE ∥ BC with AD = 2.4 cm, DB = 3.6 cm and AE = 3 cm. Find EC. If BC = 9 cm, find DE.
- BPT: AD/DB = AE/EC, so 2.4/3.6 = 3/EC.
- EC = 3 × 3.6/2.4 = 10.8/2.4 = 4.5 cm.
- AB = 2.4 + 3.6 = 6 cm. By similarity, DE/BC = AD/AB = 2.4/6 = 0.4, so DE = 0.4 × 9 = 3.6 cm.
Converse check: In triangle PQR, S on PQ and T on PR, with PS = 4 cm, SQ = 6 cm, PT = 5 cm and TR = 7.5 cm. Is ST ∥ QR? PS/SQ = 4/6 = 2/3 and PT/TR = 5/7.5 = 2/3. The ratios are equal, so yes, ST ∥ QR.
Exam tip: the most common error is writing DE/BC = AD/DB. Use AD/AB for the parallel side, and AD/DB only for the pieces of the sides.
Applications of Thales theorem
- Finding the centre of a circle. Place the corner of a set square on the circle. The two points where its edges cut the circle are the ends of a diameter (converse of Thales). Draw that diameter, repeat from another position, and the two diameters cross at the centre.
- Constructing a right angle. Draw a circle, draw any diameter, pick any point on the circle and join it to both ends: the angle there is exactly 90°. It is also how the tangent from an external point is constructed, since the radius meets the tangent at 90°.
- Heights by similar triangles. Thales is said to have measured the height of a pyramid from its shadow. A 1.5 m stick casts a 2 m shadow while a tower casts a 40 m shadow. The sun’s rays are parallel, so the triangles are similar: height = 1.5 × 40/2 = 30 m.
- Dividing a line in a given ratio. The classic compass-and-ruler construction for dividing a line segment internally in a ratio m:n draws equally spaced points on a ray and a parallel line, which is BPT at work.
References
- NCERT Mathematics – Triangles & Circles, Class 10 Chapter 6 (Theorems 6.1 and 6.2).
- Thales of Miletus, Wikipedia.
FAQs
What is Thales theorem?
Internationally, Thales’ theorem says that the angle inscribed in a semicircle is 90°: if BC is a diameter and A is any other point on the circle, ∠BAC is a right angle. In NCERT Class 10, the name is used for the Basic Proportionality Theorem: a line parallel to one side of a triangle divides the other two sides in the same ratio.
How do you prove Thales theorem for a semicircle?
Join the centre O to the point A. OA, OB and OC are radii, so triangles OAB and OAC are isosceles, with base angles α and β. The angles of triangle ABC are α, β and α + β, which add to 180°, so α + β = 90°, and that is ∠BAC.
Is the Basic Proportionality Theorem the same as Thales theorem?
In Indian textbooks, yes: NCERT calls the Basic Proportionality Theorem the Thales theorem. Elsewhere it is usually called the intercept theorem, and “Thales’ theorem” means the angle in a semicircle. Both results are attributed to Thales of Miletus.
What is the converse of Thales theorem?
For the semicircle version: if an angle BAC is 90°, then A lies on the circle with diameter BC, so the midpoint of the hypotenuse is the circumcentre of a right triangle. For BPT: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Why does the BPT proof use areas?
Triangles with the same height have areas in the ratio of their bases, which turns side ratios into area ratios. Triangles BDE and DEC have equal areas because they share base DE and lie between the parallels DE and BC, and that equality links AD/DB to AE/EC.