question_answer

In \[\Delta \,ABC,D\]and \[E\] are points \[AC\] and \[BC\] respectively such that \[DE\parallel AB\]. If \[AD=2x,\text{ }BE=2x-1,CD=x+1\] and\[CE=x-1\], then find the value of \[x\].

Answer:

\[DE\parallel AB\]

So,                   \[\frac{AB}{CD}=\frac{BE}{EC}\]                                 [By B.P.T]

\[\Rightarrow \]               \[\frac{2x}{x+1}=\frac{2x-1}{x-1}\]

\[\Rightarrow \]          \[2x(x-1)=(x+1)(2x-1)\]

\[\Rightarrow \]          \[2{{x}^{2}}-2x=2{{x}^{2}}+2x-x-1\]

\[\Rightarrow \]                 \[-2x=x-1\]

\[\Rightarrow \]                     \[1=3x\]

Or                          \[x=\frac{1}{3}\]