In Delta ABC,A-M-B,A-N-C and overline{MN} par | vedclass

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(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two s

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$1$

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(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Since $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.Substituting the given values: $\frac{3x - 1}{2x + 1} = \frac{3x + 1}{5x + 1}$.Cross-multiplying: $(3x - 1)(5x + 1) = (3x + 1)(2x + 1)$.Expanding both sides: $15x^2 + 3x - 5x - 1 = 6x^2 + 3x + 2x + 1$.$15x^2 - 2x - 1 = 6x^2 + 5x + 1$.Rearranging the terms: $9x^2 - 7x - 2 = 0$.Factoring the quadratic equation: $9x^2 - 9x + 2x - 2 = 0$.$9x(x - 1) + 2(x - 1) = 0$.$(9x + 2)(x - 1) = 0$.This gives $x = 1$ or $x = -2/9$.Since lengths must be positive,$x = 1$ is the valid solution.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 3x - 1$,$MB = 2x + 1$,$AN = 3x + 1$ and $NC = 5x + 1$,find the value of $x$.

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