If a line is drawn parallel to one side of a triangle to intersect the other two sides indistinct points the other two sides are divided in the same ratio.
Answers
Step-by-step explanation:
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.