Question

in a triangle ABC, the point D and e are on the sides CA and CB respectively such that DE parallel to AB, AD is equal to 2 x,DC is equal to X + 3 ,BE is equal to 2x - 1 and CE is equal to X then find the value of x is

Answer 1

Answer:

AD = 2x ,DC = x + 3, BE =2x - 1 ,CE = x

Using Thales Theorem

\bf\huge\frac{CD}{AD} = \frac{CE}{BE}

\bf\huge\frac{x+3}{2x} = \frac{x}{2x-1}

5x = 3

x=3/5

Answer 2

Answer:

The value of x = $\frac{3}{5}$

Step-by-step explanation:

Given,

In a triangle, ABC, the points D, and E are on the sides CA and BC respectively. and DE is parallel to AB

AD = 2x

DC = x+3

BE = 2x-1

CE = x

To find,

The value of 'x'

Solution:

Recall the Theorem

Basic Proportionality Theorem

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio

Here, In the triangle ABC, the line DE is parallel to the side AB of the triangle,

Then by basic proportionality theorem, we have,

$\frac{CD}{DA} = \frac{CE}{EB}$

Substituting the given values, we get

$\frac{x+3}{2x} = \frac{x}{2x-1}$

Cross multiplying we get,

x×2x = (x+3)(2x-1)

2x² = 2x² -x +6x-3

5x = 3

x = $\frac{3}{5}$

The value of x = $\frac{3}{5}$

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