Question

in ∆ABC D is the point on such AB and E is a point on side AC such that angle ADE =angle ABC, AD=2,BD=3 and AE =3, then what is the value of CE​

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

Step-by-step explanation:

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

.

Answer 2

Given:

In ∆ABC

angle ADE =angle ABC

AD=2

BD=3

AE =3

Find CE=?

given that angle ADE =angle ABC

By basic proportionality theorem

$\frac{AD}{BD}= \frac{AE}{CE}$

$\frac{2}{3} =\frac{3}{CE}$

$CE = \frac{3 \times 3}{2}$

$CE=\frac{9}{2}$

CE = 4.8 cm

Hence the value of CE=4.8cm