This theorem can be proved by taking a line DE such that \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) and assuming that DE is not parallel to BC.

If DE is not parallel to BC, draw a line DE' parallel to BC.
So, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}^{\prime}}{\mathrm{E}^{\prime} \mathrm{C}}\)
Therefore, \(\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{\mathrm{AE}^{\prime}}{\mathrm{E}^{\prime} \mathrm{C}}\)
Adding 1 to both sides of above, you can see that E and E' must coincide.
Let us take some examples to illustrate the use of the above theorems.

In \(\triangle\)ABC DE || AC,    (Given)
\(\frac { BD }{ DA } =\frac { BE }{ EC } \quad \quad \quad (BPT)\quad ...(i)\)
In \(\triangle\)ABE, DF || AE,    (Given)
\(\frac { BD }{ DA } =\frac { BF }{ FE } \quad \quad \quad (BPT)\quad ...(ii)\)
From (i) and (ii), we have
\(\frac { BF }{ FE } =\frac { BE }{ EC } \)
Hence proved.

DE || BC        (Given)
So, \(\begin{aligned} {\frac{\mathrm{AD}}{\mathrm{DB}}}=\frac{\mathrm{AE}}{\mathrm{EC}} \end{aligned}\)
or, \(\begin{aligned} \frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}} \\ \end{aligned}\)
or, \(\begin{aligned} \frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{\mathrm{EC}}{\mathrm{AE}}+1 \\ \end{aligned}\)
or, \(\begin{aligned} \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}} \end{aligned}\)
So, \(\begin{aligned} \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}} \end{aligned}\)

In \(\Delta\)ACB,    LM || CB         [given]
\(\Rightarrow \quad \frac{A M}{M B}=\frac{A L}{L C}\)    ....(i)
[ by basic proportionality theorem]
In \(\Delta\)ACD, LN ||CD    [given]
\(\Rightarrow \quad \frac{A N}{N D}=\frac{A L}{L C}\)    ....(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac{A M}{M B}=\frac{A N}{N D} \Rightarrow \frac{M B}{A M}=\frac{N D}{A N}\)
[on taking reciprocal of the terms]
\(\Rightarrow \quad \frac{M B}{A M}+1=\frac{N D}{A N}\) + 1  [adding 1 on both sides]
\(\begin{aligned} & \Rightarrow \quad \frac{M B+A M}{A M}=\frac{N D+A N}{A N} \\ \end{aligned}\)

\(\begin{aligned} \Rightarrow \quad \frac{A M}{A M+M B}=\frac{A N}{A N+N D} \end{aligned}\)
[on taking reciprocal of the terms]
\(\therefore \quad \frac{A M}{A B}=\frac{A N}{A D} \quad\left[\begin{array}{c} \because A D=A N+N D \\ \text { and } A B=A M+M B \end{array}\right]\)
Hence proved.

Consider \(\triangle ABC\), in which D and E are the mid-points of sides AB and AC, respectively.

\(\because \frac { AD }{ DB } =1\quad and\quad \frac { AE }{ EC } =1\Rightarrow \quad \frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\therefore DE\parallel BC\)
[by converse of basic proportionality theorem]
Hence, the line joining the mid-points of any two sides of a triangle, is parallel to the third side.
Hence proved.

Consider \(\triangle ABC\), in which D is the mid-point of AB.
Then, \(\frac { AD }{ DB } =1\)    ... (i)
Line l is drawn through D such that \(l\parallel BC\) and it meets AC at E.
By basic proportionality theorem, 
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { AE }{ EC } =1\) [from Eq. (i)]
\(\Rightarrow AE=EC\)
So, E is the mid-point of AC.
Hence, a line drawn through the mid-point of one side of a triangle parallel to another side, bisects the third side.
Hence proved.

Given, in \(\triangle ABC\), \(DE\parallel BC\)
By Thales theorem, we get
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-7 }{ 5x-3 } \)
[\(\because \)  AD = 4x - 3, DB = 3x - 1, AE = 8x - 7, EC = 5x - 3]
\(\Rightarrow \) (4x - 3)(5x - 3) = (8x - 7)(3x - 1)
\(\Rightarrow \)  20x² - 12x + 9 - 15x = 24x² - 21x - 8x + 7
\(\Rightarrow \)  4x² - 2x - 2 = 0
\(\Rightarrow \)  2x² - x - 1 = 0     [dividing both sides by 2]
\(\Rightarrow \)  2x² - 2x + x - 1 = 0   [by splitting the middle term]
\(\Rightarrow \)  2x(x - 1) + 1 (x - 1) = 0
\(\Rightarrow \)  (2x + 1) (x - 1) = 0   \(\therefore x=-\frac { 1 }{ 2 } \) or x = 1
If \(x=-\frac { 1 }{ 2 } \), then AD = \(4\times -\frac { 1 }{ 2 } -3=-5<0\)  [not possible]
Hence, x = 1 is the required value.

\(\because AB\parallel EW\)  [given]
\(\therefore \frac { DA }{ AE } =\frac { DB }{ BW } \) [by basic proportionality theorem]
\(\Rightarrow \frac { DA }{ DE-DA } =\frac { DB }{ DW-DB } \)
\(\Rightarrow \frac { 4 }{ 12-4 } =\frac { DB }{ 24-DB } \Rightarrow \frac { 4 }{ 8 } =\frac { DB }{ 24-DB } \)
\(\Rightarrow 24-DB=2DB\Rightarrow 24=3DB\Rightarrow DB=\frac { 24 }{ 3 } =\) 8 cm

In \(\triangle ABC, DE\parallel BC\)
Let EC = x cm
\(\Rightarrow \frac { AD }{ DB } =\frac { AE }{ EC } \)

[by basic proportionality theorem]
\(\Rightarrow \frac { 3 }{ 4 } =\frac { 6 }{ x } \Rightarrow \)  x = 8 cm
\(\therefore \)  EC = 8 cm

Let us join AC to intersect EF at G

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in \(\Delta\) ADC,
EG || DC (As EF || DC)
So, \(\frac{AE}{ED}=\frac{AG}{GC}\)
Similarly, from  \(\Delta\)CAB,
\(\begin{aligned} & \frac{C G}{A G}=\frac{C F}{B F} \\ \end{aligned}\)
\(\begin{aligned} & \frac{A G}{G C}=\frac{B F}{F C} \end{aligned}\)
Therefore, from (1) and (2),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)

Given, in \(\triangle ABC\), \(DE\parallel AC\)   [given]
So,  \(\frac { BE }{ EC } =\frac { BD }{ DA } \)  ... (i)
[ by basic proportionality theorem]
Also, \(DC\parallel AP\)    [given]
So, \(\frac { BC }{ CP } =\frac { BD }{ DA } \)  ... (ii)
[ by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac { BE }{ EC } =\frac { BC }{ CP } \)

In \(\triangle ABC\), \(\frac { AD }{ DB } =\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)

and \(\frac { AE }{ EC } =\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \)
So, \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Hence, \(DE\parallel BC\)
[ by converse of basic proportionallity theorem]

Given, CQ = 25 cm, CB = 30 cm, CP = 10 cm and CA = 16 cm

Here, \(\frac{C Q}{C B}=\frac{25}{30}=\frac{5}{6}\)
and \(\frac{C P}{C A}=\frac{10}{16}=\frac{5}{8} \Rightarrow \frac{C Q}{C B} \neq \frac{C P}{C A}\)
\(\Rightarrow \frac{C B}{C Q} \neq \frac{C A}{C P} \Rightarrow \frac{C B}{C Q}-1 \neq \frac{C A}{C P}-1\)
\(\Rightarrow \frac{C B-C Q}{C Q} \neq \frac{C A-C P}{C P}\)
\(\Rightarrow \frac{Q B}{C Q} \neq \frac{P A}{C P} \text { or } \frac{C Q}{Q B} \neq \frac{C P}{P A}\)
Then, by converse of basic proportionality theorem, PQ is not parallel to AB.

(i) Examples of similar figures:
(a) All squared
(b) All regular hexagons
(ii) Examples of non-similar figures:
(a) Two isosceles triangles of different angle measures
(b) Two rhombus of different angle measures.

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
\( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \)
\(\frac{P F}{P R}=\frac{0.36}{2.56}=\frac{9}{64} \)
Hence \(\frac{P E}{P Q}=\frac{P F}{P R}\)
Therefore EF is parallel to QR

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
\( \frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9} \)
\(\frac{P F}{F R}=\frac{8}{9} \)
\(\frac{P F}{F R}=\frac{P E}{E Q} \)
Therefore, EF is parallel to QR

Let AD = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\( \frac{A D}{D B}=\frac{A E}{E C} \)
\(\frac{x}{7.2}=\frac{1.8}{5.4} \)
\(x=\frac{1.8 \times 7.2}{5.4}\)
x = 2.4
∴ AD = 2.4 cm

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{1.5}{3}=\frac{1}{x}\)
\(x=\frac{3 \times 1}{1.5}\)
x =  2
∴ EC = 2 cm

2.7 cm

x = 4

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

(i) (a) Pair of the equilateral triangles are similar figures.
(b) Pair of the squares are similar figures.
(ii) (a) A triangle and a quadrilateral form a pair of non-similar figures.
(b) A square and a circle form a pair of non-similar figures.

(i) AIl circles are similar because all circles have same shape but size can vary.
(ii) All squares are similar because all squares have same shape but size can vary.
(iii) All equilateral triangles are similar because all equilateral triangle have same shape but size can vary.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are cqual and
(b) their corresponding sides are proportional.