Theorem 1: Basic Proportionality Theorem (BPT) or Thales theorem,
Theorem 2: Converse of Basic Proportionality Theorem,
Theorem 3: Angle Bisector Theorem,
Theorem 4: Converse of Angle Bisector Theorem

**Thales Theorem and Angle Bisector Theorem**

Thales, (640 - 540 BC (BCE)) the most famous Greek mathematician
and philosopher lived around seventh century BC (BCE). He possessed knowledge
to the extent that he became the first of seven sages of Greece. Thales was the
first man to announce that any idea that emerged should be tested
scientifically and only then it can be accepted. In this aspect, he did great
investigations in mathematics and astronomy and discovered many concepts. He
was credited for providing first proof in mathematics, which today is called by
the name “Basic Proportionality Theorem”. It is also called “Thales Theorem”
named after its discoverer.

The discovery of the Thales theorem itself is a very interesting
story. When Thales travelled to Egypt, he was challenged by Egyptians to
determine the height of one of several magnificent pyramids that they had
constructed. Thales accepted the challenge and used similarity of triangles to
determine the same successfully, another triumphant application of Geometry.
Since *X*_{0}, *X*_{1} and *H*_{0} are
known, we can determine the height H_{1} of the pyramid.

To understand the basic proportionality theorem or Thales theorem, let us do the following
activity.

A straight line drawn parallel to a side of triangle intersecting
the other two sides, divides the sides in the same ratio.

**Given: **In Δ*ABC* , *D* is a point on *AB* and *E* is a point on *AC*.

*AD/DB* =
*AE/EC*

Draw a line *DE || BC*

**Corollary **

If in Δ*ABC* , a straight line *DE* parallel to *BC*,
intersects *AB* at *D* and *AC* at *E*, then

In Δ*ABC*, *DE* || *BC*.

therefore, => *AD/DB* = *AE/EC *(by Basic
Proportionality Theorem)

Is the converse of Basic Proportionality Theorem also true? To examine let us do the following illustration.

Draw an angle *XAY* on your notebook as shown in Fig.4.31 and
on ray *AX*, mark points *B*_{1} , *B*_{2} , *B*
_{3} , *B*_{4} and *B* such that *AB*_{1}
= *B*_{1}*B*_{2} = *B*_{2} *B*_{3}
= *B* _{3} *B*_{4} = *B* _{4} *B* =
1 cm.

Similarly on ray *AY*, mark points *C*_{1}, *C*_{2}
, *C*_{3} ,*C*_{4} and *C*, such that *AC*_{1}=*C*_{1}*C*_{2}*
*=* C*_{2}*C*_{3}* *=* C*_{3}*C*_{4}*
*=* C*_{4}*C *= 2 cm, Join* B*_{1}* C*_{1}*
*and* BC *

Observe that

Similarly joining *B*_{2}*C*_{2},* B*_{3}*C*_{3}*
*and* B*_{4}*C*_{4}* *you see that

From this we observe that if a line divides two sides of a
triangle in the same ratio, then the line is parallel to the third side.

Therefore, we obtain the following theorem called converse of the Thales theorem.

**Theorem 2:**

If a straight line divides any two sides of a triangle in the same
ratio, then the line must be parallel to the third side.

**Given :** In *ABC*,

**To prove :** *DE || BC *

**Construction** Draw *BF || DE*

The internal bisector of an angle of a triangle divides the
opposite side internally in the ratio of the corresponding sides containing the
angle.

**Given :**** **** **In ΔABC, AD is the internal
bisector

**To prove :**

**Construction :** Draw a line through *C* parallel to *AB*. Extend *AD* to
meet line through *C* at *E*

**Theorem 4:**

If a straight line through one vertex of a triangle divides the
opposite side internally in the ratio of the other two sides, then the line
bisects the angle internally at the vertex.

**Given : **ABC is a triangle. *AD* divides *BC* in the ratio of the
sides containing the angles ∠*A* to meet *BC* at *D*.

That is

**To prove :** AD bisects ∠*A *i.e. ∠1 = ∠2

**Construction :** Draw *CE || DA* . Extend *BA* to meet at *E*.

**Example 4.12**

In** **Δ*ABC*** **,** **if** ***DE || BC*** **, *AD *=* x*,* DB *=* x *−*
*2,* *and* EC *=* x *−1* *then find the lengths of
the sides *AB* and *AC.*

*Solution*

In Δ*ABC* we have *DE || BC *.

By Thales theorem, we have

*x/ x*-2 = (*x* + 2)/(*x* - 1) gives *x*
(*x* − 1) = (*x* − 2)(*x* + 2)

Hence, *x* ^{2} − *x* = *x*^{2} – 4 so, *x *=* *4

When *x* = 4 , *AD* = 4 , *DB* = *x* − 2 = 2 ,
*AE* = *x* + 2 = 6 , *EC* = *x – 1 =3*

Hence, *AB* = *AD* + *DB* = 4 + 2 = 6 , *AC* =
*AE* + *EC* = 6 + 3 = 9 .

Therefore, *AB* = 6, *AC* = 9.

*D*** **and** ***E*** **are respectively the points on the sides** ***AB*** **and** ***AC*** **of a** **Δ*ABC*** **such that *AB*=5.6
cm, AD=1.4 cm, *AC*=7.2 cm and *AE* = 1.8 cm, show that *DE || BC*.

We have* **AB*** **=

*BD *=* AB *−* AD *= 5.6–1.4 = 4.2* *cm

and *EC* = *AC* –*AE* = 7.2–1.8 = 5.4 cm.

Therefore, by converse of Basic Proportionality Theorem, we have *DE* is
parallel to *BC.*

Hence proved.

In the Fig.4.38,** ***DE || AC *and* DC || AP *. Prove that .

Solution In ΔBPA, we have DC || AP . By Basic Proportionality
Theorem,

we have

In Δ*BCA*, we have *DE || AC* . By Basic Proportionality
Theorem, we have

From (1) and (2) we get, *BE/EC* = *BC/CP => *. Hence proved.

In the Fig.4.39,** ***AD*** **is the bisector of ∠*A* . If *BD* = 4 cm, *DC *=* *3* *cm and* AB *=*
*6* *cm, find* AC.*

In Δ*ABC* , *AD* is the bisector of ∠*A *

Therefore by Angle Bisector Theorem

4/3 = 6/ AC

gives 4AC = 8 . Hence, AC = 9/2 =4.5 cm

In the Fig. 4.40,** ***AD*** **is the bisector of** **∠*BAC*** **, if *AB* = 10 cm, *AC* = 14 cm and BC = 6 cm. Find *BD*
and *DC*.

Let *BD* = *x* cm, then *DC* = (6–*x*)cm

AD is the bisector of ∠*A*

Therefore by Angle Bisector Theorem

Therefore, *BD* = 2.5 cm, *DC *=* *6* *−*
x *=* *6* *−* *2.5* *= 3.5 cm

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