In the past, a famous Greek mathematician named Thales came up with a fact relating 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 understand the above-stated concept introduced in similar triangles better.
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.
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 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.
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.
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.
XP/PY = XQ/QZ (By Basic Proportionality Theorem)
⟹ 6cm/2cm = 7.5cm/QZ
⟹ QZ = (7.5x2)/6
⟹ QZ = 2.5 cm.
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.
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.
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.
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