Question

show how to find points D and E on the side AB,AC of ∆ABC such that DE||BC and DE=BD​

Answer 1

Answer:

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that AD=4x−38, BD=3x−1, AE=8x−7 and CE=5x−3. Let AC=x

Using the basic proportionality theorem, we have

BD

AD

=

AC

AE

⇒

3x−1

4x−3

=

x

8x−5

⇒x(4x−3)=(3x−1)(8x−5)

⇒4x

2

−3x=3x(8x−5)−1(8x−5)

⇒4x

2

−3x=24x

2

−15x−8x+5

⇒4x

2

−3x=24x

2

−23x+5

⇒24x

2

−23x+5−4x

2

+3x=0

⇒20x

2

−20x+5=0

⇒5(4x

2

−4x+1)=0

⇒4x

2

−4x+1=0

⇒(2x)

2

−(2×2x×1)x+1

2

=0(∵(a−b)

2

=a

2

+b

2

−2ab)

⇒(2x−1)

2

=0

⇒(2x−1)=0

⇒2x=1

⇒x=

2

1

Hence, x=

2

1

.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

your answer is not clear