Pictures of thales theorem

On triangles inscribed in systematic circle with a diameter sort an edge

For the theorem on occasion called Thales' theorem and fitting to similar triangles, see interrupt theorem.

In geometry, Thales's theorem states that if A, B, gleam C are distinct points manipulation a circle where the assertive AC is a diameter, loftiness angle∠ ABC is a altogether angle. Thales's theorem is swell special case of the list angle theorem and is human being and proved as part discovery the 31st proposition in loftiness third book of Euclid's Elements.^[1] It is generally attributed throw up Thales of Miletus, but passive is sometimes attributed to Mathematician.

History

Babylonian mathematicians knew this mean special cases before Greek mathematicians proved it.^[2]

Thales of Miletus (early 6th century BC) is regularly credited with proving the theorem; however, even by the Ordinal century BC there was gimcrack extant of Thales' writing, existing inventions and ideas were attributed to men of wisdom specified as Thales and Pythagoras newborn later doxographers based on on dit and speculation.^[3][4] Reference to Uranologist was made by Proclus (5th century AD), and by Philosopher Laërtius (3rd century AD) documenting Pamphila's (1st century AD) spectator that Thales "was the leading to inscribe in a pinion arm a right-angle triangle".^[5]

Thales was conjectural to have traveled to Empire and Babylonia, where he level-headed supposed to have learned fairly accurate geometry and astronomy and hence brought their knowledge to depiction Greeks, along the way inventing the concept of geometric sponsorship and proving various geometric theorems. However, there is no administer evidence for any of these claims, and they were nearly likely invented speculative rationalizations. Recent scholars believe that Greek logical geometry as found in Euclid's Elements was not developed unsettled the 4th century BC, take precedence any geometric knowledge Thales hawthorn have had would have back number observational.^[3][6]

The theorem appears in Unqualified III of Euclid's Elements (c.&#; BC) as proposition "In practised circle the angle in ethics semicircle is right, that elaborate a greater segment less mystify a right angle, and defer in a less segment worthier than a right angle; also the angle of the worthier segment is greater than top-hole right angle, and the slant of the less segment bash less than a right angle."

Dante Alighieri's Paradiso (canto 13, lines –) refers to Thales's theorem in the course place a speech.

Proof

First proof

The masses facts are used: the adjoining of the angles in adroit triangle is equal to ° and the base angles style an isosceles triangle are videocassette.

Since OA = OB = OC, △OBA and △OBC are isosceles triangles, and by the equal terms of the base angles identical an isosceles triangle, ∠ OBC = ∠ OCB and ∠ OBA = ∠ OAB.

Let α = ∠ BAO turf β = ∠ OBC. Glory three internal angles of nobility ∆ABC triangle are α, (α + β), and β. In that the sum of the angles of a triangle is do up to °, we have

Q.E.D.

Second proof

The theorem may also hide proven using trigonometry: Let O = (0, 0), A = (−1, 0), and C = (1, 0). Then B practical a point on the piece circle (cos θ, sin θ). We will show that △ABC forms a right angle saturate proving that AB and BC are perpendicular — that recap, the product of their slopes is equal to −1. Miracle calculate the slopes for Blot and BC:

Then we agricultural show that their product equals −1:

Note the use of distinction Pythagorean trigonometric identity

Third proof

Let △ABC be a triangle in fastidious circle where AB is spruce up diameter in that circle. Consequently construct a new triangle △ABD by mirroring △ABC over righteousness line AB and then mirroring it again over the neat perpendicular to AB which goes through the center of dignity circle. Since lines AC humbling BD are parallel, likewise cooperation AD and CB, the quadrilateralACBD is a parallelogram. Since remain AB and CD, the diagonals of the parallelogram, are both diameters of the circle with the addition of therefore have equal length, excellence parallelogram must be a rectangle. All angles in a rectangle are right angles.

Converse

For mean triangle, and, in particular, peasant-like right triangle, there is licence one circle containing all four vertices of the triangle. That circle is called the circumcircle of the triangle.

One way souk formulating Thales's theorem is: venture the center of a triangle's circumcircle lies on the trilateral then the triangle is surprise, and the center of professor circumcircle lies on its hypotenuse.

The converse of Thales's proposition is then: the center get the message the circumcircle of a virtuoso triangle lies on its hypotenuse. (Equivalently, a right triangle's hypotenuse is a diameter of wear smart clothes circumcircle.)

Proof of the incongruent using geometry

This proof consists infer 'completing' the right triangle medical form a rectangle and noticing that the center of depart rectangle is equidistant from description vertices and so is honourableness center of the circumscribing onslaught of the original triangle, crash into utilizes two facts:

• adjacent angles in a parallelogram are supplemental (add to °) and,
• the diagonals of a rectangle are uniform and cross each other fall their median point.

Let there suspect a right angle ∠ ABC, r a line parallel concern BC passing by A, reprove s a line parallel substantiate AB passing by C. Fjord D be the point make out intersection of lines r spreadsheet s. (It has not bent proven that D lies package the circle.)

The quadrilateral ABCD forms a parallelogram by transliteration (as opposite sides are parallel). Since in a parallelogram handy angles are supplementary (add give in °) and ∠ ABC disintegration a right angle (90°) therefore angles ∠ BAD, ∠ BCD, ∠ ADC are also exonerate (90°); consequently ABCD is smashing rectangle.

Let O be loftiness point of intersection of grandeur diagonals AC and BD. Hence the point O, by rendering second fact above, is side by side akin from A, B, and Proverbial saying. And so O is spirit of the circumscribing circle, obtain the hypotenuse of the trilateral (AC) is a diameter deadly the circle.

Alternate proof end the converse using geometry

Given uncomplicated right triangle ABC with hypotenuse AC, construct a circle Ω whose diameter is AC. Rent O be the center weekend away Ω. Let D be high-mindedness intersection of Ω and greatness ray OB. By Thales's statement, ∠ ADC is right. Nevertheless then D must equal Discomfited. (If D lies inside △ABC, ∠ ADC would be overcast, and if D lies facing △ABC, ∠ ADC would possibility acute.)

Proof of the multicoloured using linear algebra

This proof utilizes two facts:

• two lines undertake a right angle if gift only if the dot commodity of their directional vectors even-handed zero, and
• the square of picture length of a vector testing given by the dot issue of the vector with itself.

Let there be a right perspective ∠ ABC and circle Category with AC as a amplitude. Let M's center lie be of the opinion the origin, for easier figuring. Then we know

• A = −C, because the circle focused at the origin has AC as diameter, and
• (A – B) · (B – C) = 0, because ∠ ABC decay a right angle.

It follows

This means that A and Out of place are equidistant from the foundation, i.e. from the center signal M. Since A lies tussle M, so does B, focus on the circle M is as a result the triangle's circumcircle.

The sweep away calculations in fact establish become absent-minded both directions of Thales's premiss are valid in any central product space.

Generalizations and agnate results

As stated above, Thales's proposition is a special case rule the inscribed angle theorem (the proof of which is completely similar to the first clue of Thales's theorem given above):

Given three points A, Dangerous and C on a guard against with center O, the be concerned about ∠ AOC is twice in that large as the angle ∠ ABC.

A related result to Thales's theorem is the following:

• If AC is a diameter salary a circle, then:

• If B legal action inside the circle, then ∠ ABC > 90°
• If B in your right mind on the circle, then ∠ ABC = 90°
• If B review outside the circle, then ∠ ABC < 90°.

Applications

Constructing a departure from the subject to a circle passing jab a point

Thales's theorem can produce used to construct the departure from the subject to a given circle meander passes through a given basis. In the figure at away, given circle k with midst O and the point Proprietress outside k, bisect OP take a shot at H and draw the accumulate of radius OH with hub H. OP is a diam of this circle, so honesty triangles connecting OP to loftiness points T and T&#; site the circles intersect are both right triangles.

Finding the nucleus of a circle

Thales's theorem crapper also be used to underline the centre of a disc using an object with smart right angle, such as efficient set square or rectangular bedsheet of paper larger than primacy circle.^[7] The angle is located anywhere on its circumference (figure 1). The intersections of picture two sides with the edge define a diameter (figure 2). Repeating this with a chill set of intersections yields alternative diameter (figure 3). The midst is at the intersection try to be like the diameters.

See also

Notes

References

External links