Using the Basic Proportionality Theorem, show that if a line intersecting two sides of a triangle
and parallel to the third side cuts segments of the intersecting sides such that segments in the
same half plane is proportional to their corresponding sides.
Answers
Answer:
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.
Solved Examples
Example 1: Consider
Δ
A
B
C
, in which segment
D
E
parallel to
B
C
is drawn from
D
on
A
B
to
E
on
A
C
, as shown below:
Line parallel to base of triangle
Show that
Δ
A
D
E
is similar to
Δ
A
B
C
.
Solution: Recall that two triangles are said to be similar if they are equi-angular (corresponding angles are equal), and corresponding sides are proportional. It is obvious that
Δ
A
D
E
and
Δ
A
B
C
are equi-angular, since:
∠
A
=
∠
A
(common angle)
∠
A
D
E
=
∠
A
B
C
(corresponding angles)
∠
A
E
D
=
∠
A
C
B
(corresponding angles)
Now, we will show that the corresponding sides are proportional. Using the BPT, we have:
A
D
D
B
=
A
E
E
C
⇒
D
B
A
D
=
E
C
A
E
⇒
D
B
A
D
+
1
=
E
C
A
E
+
1
⇒
D
B
+
A
D
A
D
=
E
C
+
A
E
A
E
⇒
A
B
A
D
=
A
C
A
E
⇒
A
D
A
B
=
A
E
A
C
We have thus shown two pairs of sides to be proportional. All that remains to be shown that the ratios above are also equal to the ratio of the third pair, that is, to
D
E
B
C
.
To prove that, draw segment
E
F
parallel to
A
B
, as shown:
Similar triangles are equi-angular
Since
E
F
∥
A
B
, the BPT tells us that:
C
E
E
A
=
C
F
F
B
⇒
C
E
E
A
+
1
=
C
F
F
B
+
1
⇒
C
E
+
E
A
E
A
=
C
F
+
F
B
F
B
⇒
A
C
A
E
=
B
C
F
B
But
B
F
is equal to
D
E
(since
D
E
F
B
is a parallelogram), and so:
A
C
A
E
=
B
C
D
E
⇒
A
E
A
C
=
D
E
B
C
Thus, we finally have:
A
D
D
B
=
A
E
A
C
=
D
E
B
C
This proves that all the three pairs of sides are proportional, which means that the two triangles are similar.
⇒
Δ
A
D
E
∼
Δ
A
B
C
✍Note: The result we have discussed here is extremely important and will be used frequently, so it is imperative that you go through the discussion again if you have not followed it completely.
Answer:
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