simple question that what is basic proportionality theorem it for (8-10) marks
Answers
Introduction:
A famous Greek mathematician Thales gave an important truth relating two equi-angular triangles,i.e., "The ratio of any two corresponding sides in two equiangular triangles is always the same". Thales used a result called the Basic Proportionality Theorem for the same. Before discussing other criterions and theorems of similar triangles, it is important to understand this very fundamental theorem related to triangles: the Basic Proportionality Theorem or BPT Theorem. This theorem is a key to understanding the concept of similarity better.
Basic Proportionality Theorem:
Basic Proportionality Theorem states that "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio".
In the following figure, segment
D
E
is parallel to the side
B
C
of
Δ
A
B
C
. Note how
D
E
divides
A
B
and
A
C
in the same ratio:
Intersecting triangle proportionally
Proof of Basic Proportionality Theorem:
Given:
Δ
A
B
C
D
E
∥
B
C
To prove:
A
D
D
B
=
A
E
E
C
Construction:
Join
B
E
and
C
D
Draw
D
P
⊥
A
C
Draw
E
Q
⊥
A
B
Intersecting triangle proportionally
Proof: Consider
Δ
A
E
D
. If you have to calculate the area of this triangle, you can take
A
D
to be the base, and
E
Q
to be the altitude, so that:
a
r
(
Δ
A
E
D
)
=
1
2
×
A
D
×
E
Q
Now, consider
Δ
D
E
B
. To calculate the area of this triangle, you can take
D
B
to be the base, and
E
Q
(again) to be the altitude (perpendicular from the opposite vertex
E
).
Thus,
a
r
(
Δ
D
E
B
)
=
1
2
×
D
B
×
E
Q
Next, consider the ratio of these two areas you have calculated:
a
r
(
Δ
A
E
D
)
a
r
(
Δ
D
E
B
)
=
1
2
×
A
D
×
E
Q
1
2
×
D
B
×
E
Q
=
A
D
D
B
In an exactly analogous manner, you can evaluate the ratio of areas of
Δ
A
E
D
and
Δ
E
D
C
:
a
r
(
Δ
A
E
D
)
a
r
(
Δ
E
D
C
)
=
1
2
×
A
E
×
D
P
1
2
×
E
C
×
D
P
=
A
E
E
C
Finally, We know that "Two triangles on the same base and between the same parallels are equal in area". Here,
Δ
D
E
B
and
Δ
E
D
C
are on the same base
D
E
and between the same parallels –
D
E
∥
B
C
.
⇒
a
r
(
Δ
D
E
B
)
=
a
r
(
Δ
E
D
C
)
Considering above results, we can note,
a
r
(
Δ
A
E
D
)
a
r
(
Δ
D
E
B
)
=
a
r
(
Δ
A
E
D
)
a
r
(
Δ
E
D
C
)
⇒
A
D
D
B
=
A
E
E
C
This completes our proof of the fact that
D
E
divides
A
B
and
A
C
in the same ratio.
Hence Proved.
✍Note:
D
E
is parallel to
B
C
was crucial to the proof. Without this, the areas of
Δ
D
E
B
and
Δ
E
D
C
would not have been the same, and hence the two ratios would have been different.
Can we say that the converse of the Basic Proportionality Theorem (BPT) will hold? That is if, in a triangle, a line segment divides two sides in the same ratio, will it be parallel to the third side? The answer is yes. Let’s prove this result.
The converse of BPT
The converse of BPT states that "In a triangle, if a line segment intersecting two sides and divides them in the same ratio, then it will be parallel to the third side".
Proof of converse of BPT:
Consider the following figure,
It is given that
A
D
D
B
=
A
E
E
C
.
Now, Suppose that
D
E
is not parallel to
B
C
. Draw segment
D
F
through
D
which is parallel to
B
C
, as shown:
Converse of Basic Proportionality Theorem
Using the BPT, we see that
D
F
should divide
A
B
and
A
C
in the same ratio. Thus, we should have:
A
D
D
B
=
A
F
F
C
But it is already given to us that:
A
D
D
B
=
A
E
E
C
This means that:
A
E
E
C
=
A
F
F
C
Adding 1 to both sides, we have,
A
E
E
C
+
1
=
A
F
F
C
+
1
⇒
A
E
+
E
C
E
C
=
A
F
+
F
C
F
C
⇒
A
C
E
C
=
A
C
F
C
⇒
E
C
=
F
C
This cannot happen if
E
and
F
are different points, so they must coincide. Thus, we can conclude that
D
E
is parallel to
B
C
, hence completing the proof of the converse of the BPT.
Hence Proved.
Answer:
1) suppose if there is a simple triangle ABC and a seg PQ passing through it which is parallel to the traingles base then-
AP/PB=AQ/QC