Basic Proportionality Theorem (BPT)

In the past, a famous Greek mathematician named Thales came up with a fact relating to two equiangular triangles. The crucial truth provided by Thales states that the ratio of any two corresponding sides of two equiangular triangles is always the same. Based on this concept, the mathematician later gave a theorem known as Basic Proportionality Theorem (BPT) or Thales theorem. Here, we will state and prove BPT, which is the key to understanding the above-stated concept introduced in similar triangles better. 

Thales Theorem Statement

Thales or Basic Proportionality Theorem States: If a line is drawn parallel to one side of a triangle to intersect its other two sides in distinct points, then the other two sides are divided in the same ratio. 

Applying the statement on the triangle ABC (given below), we can say that if the line segment DE is drawn parallel to the side BC of triangle ABC to intersect sides AB and AC respectively, then DE divides AB and AC in the same ratio. 

Proof of the Basic Proportionality Theorem

Given, 

1. Triangle ABC

2. DE ∥ BC 

To Prove: According to the BPT stated above, we need to prove:

AD/DB = AE/EC

Construction: From vertex B, draw a line meeting the side AC of triangle ABC at E to form a line BE. Now, from E draw a perpendicular EN to the side AB. Similarly, join the vertex C to D to form a line CD and then, draw a perpendicular DM to the side AC, as shown in the figure.

Proof: Consider triangle ADE and recall the formula for the area of a triangle, which is 1/2 x base x height. Now, use the formula to calculate the area of triangle ADE, where AD is the base and EN is the altitude, i.e., height as EN ⊥ AD. 

So, ar(ΔADE) =1/2 × AD × EN 

Similarly, ar(ΔDEB) =1/2 × DB × EN

Now, determine the ratio of the two areas calculated above:

ar(ΔADE)/ ar(ΔDEB) = (1/2 × AD × EN )/ (1/2 × DB × EN) = AD/DB…(1)

In an analogous manner, calculate the areas of ΔAED and ΔEDC

ar(ΔAED) = 1/2 × AE × DM (because AE is the base and DM is the height)

ar(ΔEDC) = 1/2 × EC × DM (because EC is the base and DM is the height)

Now, by calculating the ratio of areas of triangles ΔAED and ΔEDC, we get:

ar(ΔAED) /ar(ΔEDC) = (1/2 × AE × DM)/ (1/2 × EC × DM) = AE/EC…(2)

Finally, recall the property of triangles that says - two triangles on the same base and between the same parallel lines are equal in area. 

And, by using this property for ΔDEB and ΔEDC that are on the same base DE and between the same parallel lines DE∥BC, you can say that:

⇒ ar(ΔDEB) = ar(ΔEDC)…(3)

Considering the above equations and results, you can conclude that:

ar(ΔADE)/ar(ΔDEB) = ar(ΔAED)/ar(ΔEDC)

⇒ AD/DB = AE/EC Hence proved.

With the proof of the basic proportionality theorem, we arrive at conclusions like:

• DE divides the sides AB and AC in the same ratio. 

• If D and E are the mid-points of the sides AB and AC respectively, then DE || BC. 

• From this, the converse of the mid-point theorem also appears true. It states that if a line drawn from the midpoint of one side of a triangle is parallel to another side, then it bisects the third side. 

Hence, BPT or Thales Theorem is proved.

Note that the fact DE || BC was crucial for the above proof. Without this, ar(ΔDEB) and ar(ΔEDC) would not have been equal, and as a result, the two ratios – AD/DB and AE/EC would have been different.  

So, we can say that the converse of the basic proportionality theorem is also important, and let’s prove it.

The Converse of Basic Proportionality Theorem

Statement: It states that if a line intersects the two sides of a triangle such that it divides them in the same ratio, then the line will be parallel to the third side.

Proof: Given, AD/DB = AE/EC

Suppose that DE is not parallel to BC. Now, construct a line DF parallel to BC as shown in the above figure.

Using BPT, you see that DF divides AB and AC in the same ratio.

So, AD/DB = AF/FC                              

But, AD/DB = AE/EC (given)

It means, AE/EC = AF/FC 

Now, adding 1 to both the sides, you get:

(AE/EC) + 1 = (AF/FC) +1

⇒ (AE + EC)/EC = (AF + FC)/FC

⇒ AC/EC = AC/FC

⇒ EC = FC

This result cannot be obtained if E and F are different points. In other words, this is possible only when E and F coincide. So, you can conclude that DE || BC. Hence, the converse of the BPT is proved.

History

In calculation, Thales' hypothesis expresses that if A, B, and C are unmistakably focused on a circle where the line AC is a breadth, the point ABC is a right point. Thales' hypothesis is an extraordinary instance of the recorded point hypothesis and is referenced and demonstrated as a component of the 31st suggestion in the third book of Euclid's Elements. It is by and large credited to Thales of Miletus, yet it is once in a while ascribed to Pythagoras. 

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There isn't anything surviving from the composition of Thales. Work done in old Greece would in general be ascribed to men of shrewdness without regard to every one of the people engaged with specific scholarly developments; this is valid for Pythagoras particularly. Attribution would in general happen sometime in the future. Reference to Thales was made by Proclus, and by Diogenes Laërtius recording Pamphila's explanation that Thales "was quick to write in a circle a right-point triangle".

Indian and Babylonian mathematicians knew this for exceptional cases before Thales demonstrated it. It is accepted that Thales discovered that a point recorded in a crescent is a right point during his movements to Babylon. The hypothesis is named after Thales since he was said by old sources to have been quick to demonstrate the hypothesis, utilising his own outcomes that the base points of an isosceles triangle are equivalent, and that the amount of points of a triangle is equivalent to 180°.

The Converse of Basic Proportionality Theorem

For any triangle, and, specifically, any right triangle, there is by and large one circle containing every one of the three vertices of the triangle. (Sketch of confirmation. The locus of focuses equidistant from two given focuses is a straight line that is known as the opposite bisector of the line fragment interfacing the focuses. The opposite bisectors of any different sides of a triangle meet at precisely one point. This point should be equidistant from the vertices of the triangle.) This circle is known as the circumcircle of the triangle.

One method of detailing Thales' hypothesis is: assuming the focal point of a triangle's circumcircle lies on the triangle, the triangle is correct, and the focal point of its circumcircle lies on its hypotenuse.

The opposite of Thales' hypothesis is then, at that point: the focal point of the circumcircle of a right triangle lies on its hypotenuse. (Comparably, a right triangle's hypotenuse is the width of its circumcircle.)

Proof of the Converse Using Geometry

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This evidence comprises of 'finishing' the right triangle to shape a square shape and seeing that the focal point of that square shape is equidistant from the vertices as is the focal point of the delineating circle of the first triangle, it uses two realities: contiguous points in a parallelogram are strengthening (add to 180°) and, the diagonals of a square shape are equivalent and cross each other in their middle point.

Let there be a right point ∠ABC, r a line corresponding to BC passing by An and s a line corresponding to AB passing by C. Leave D alone the mark of convergence of lines r and s (Note that it has not been demonstrated that D lies on the circle). The quadrilateral ABCD structures a parallelogram by development (as inverse sides are equal). Since in a parallelogram contiguous points are strengthening (add to 180°) and ∠ABC is a right point (90°) then, at that point, points ∠BAD, ∠BCD, and ∠ADC are likewise correct (90°); subsequently ABCD is a square shape. Leave O alone the mark of the crossing point of the diagonals AC and BD. Then, at that point, the point O, continuously true above, is equidistant from A, B, and C. Thus O is the focus of the delineating circle, and the hypotenuse of the triangle (AC) is a measurement of the circle.

Substitute Confirmation of the Opposite Utilising Maths

Given a right triangle ABC with hypotenuse AC, develop a circle Ω whose measurement is AC. Leave O alone the focal point of Ω. Leave D alone the convergence of Ω and the beam OB. By Thales' hypothesis, ∠ADC is correct. However, at that point D should rise to B. (On the off chance that D lies inside ∆ABC, ∠ADC would be harsh and assuming D lies outside ∆ABC, ∠ADC would be intense.)

Verification of the Opposite Utilising Straight Polynomial Mathematics

This verification uses two realities:

Two lines structure a right point if and provided that the speck result of their directional vectors is zero, and the square of the length of a vector is given by the dab result of the vector with itself.

May there be a right point ∠ABC and circle M with AC as a breadth. Allow M's to fixate on the beginning, for simpler computation. Then, at that point, we know

A = − C, on grounds that the circle focused at the beginning, has AC as width, and

(A − B) · (B − C) = 0, on the grounds that ∠ABC is the right point.

It follows

0 = (A − B) · (B − C) = (A − B) · (B + A) = |A|2 − |B|2.

Hence,

|A| = |B|.

This implies that An and B are equidistant from the beginning, for example from the focal point of M. Since A lies on M, B does as well, and the circle M is subsequently the triangle's circumcircle.

The above estimations indeed buildup that the two headings of Thales' hypothesis are substantial in any internal item space.

Speculations and Related Outcomes

Thales' hypothesis is a unique instance of the accompanying hypothesis:

Given three focuses A, B, and C on a circle with focus O, the point ∠AOC is twice pretty much as extensive as the point ∠ABC.

See recorded point, the verification of this hypothesis is very like the evidence of Thales' hypothesis given previously.

A connected outcome to Thales' hypothesis is the accompanying:

Assuming that AC is a measurement of a circle, then, at that point:

Assuming that B is inside the circle, then, at that point, ∠ABC > 90°

Assuming that B is on the circle, then, at that point, ∠ABC = 90°

Assuming that B is outside the circle, then, at that point, ∠ABC < 90°.

Building a Digression Utilising Thales' Hypothesis

Thales' hypothesis can be utilised to build the digression to a given circle that goes through a given point. In the figure at right, given circle k with focus O and the point P outside k, divide OP at H and draw the circle of span OH with focus H. Operation is a measurement of this circle, so the triangles associating OP to the focuses T and T′ where the circles meet are both right triangles.

Thales' hypothesis can likewise be utilised to observe the focal point of a circle utilising an item with a right point, for example, a set square or rectangular piece of paper bigger than the circle. The point is put anywhere on its circuit. The crossing points of the different sides with the outline characterise a width. Rehashing this with an alternate arrangement of crossing points yields another width. The middle is at the crossing point of the breadths.

Basic Proportionality Theorem – Solved Examples

1. In a given ∆XYZ, the line PQ meeting the sides XY and XZ at P and Q respectively is parallel to YZ. If XP = 6 cm, XQ = 7.5 cm, PY = 2 cm, then find QZ.                              

Solution: 

In ∆XYZ, 

XP/PY = XQ/QZ (By Basic Proportionality Theorem)

⟹ 6cm/2cm = 7.5cm/QZ

⟹ QZ = (7.5x2)/6

⟹ QZ = 2.5 cm.

2. If in a ∆ABC, a line l which is parallel to side BC intersects AB and AC at points D and E respectively, then prove that AD/AB = AE/AC.

Solution: 

Given: Line l is parallel to side BC, i.e., DE // BC 

To prove: AD/AB = AE/AC

Proof: DE // BC (given)

So, AD/DB = AE/EC (By Basic Proportionality Theorem)          … (1) 

Now, by interchanging the ratios in equation (1) and then adding 1 on both sides, we get

(DB/AD)  + 1 = (EC/AE) + 1

(DB + AD)/AD = (EC + AE)/AE

AB/ AD = AC/ AE                                                                            …(2)

Finally, after interchanging the ratios obtained in equation (2), we get

AD/AB = AE/AC Hence proved.

3. In the figure given below DE ||BC. If AD = x cm, DB = x-2 cm, AE = x +2, and EC = x-1 cm, then find the value of x.                                          

Solution:

In the ∆ABC, DE ||BC (given)

So, AD /DB = AE /EC (by basic proportionality theorem)   … (1)

Now, putting the value of  AD, DB, AE, and EC in equation (1), we get:

⇒ x /(x - 2) = (x + 2) /(x - 1)

⇒ x (x - 1) = (x - 2)(x + 2)

⇒ x2 - x = x2 – 4

⇒ -x = -4

⇒ x = 4. 

Hence, the value of x = 4

For the hypothesis at times called Thales' hypothesis and relating to comparative triangles, see catch hypothesis.

Thales' hypothesis: on the off chance that AC is width and B is a point on the measurement's circle, the point ABC is a right point.