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

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(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 $\frac{AM}{MB} = \frac{AN}{NC}$.Given $AM = x+3$ and $AB = 2x$,then $MB = AB - AM = 2x - (x+3) = x-3$.Given $AN = x+5$ and $AC = 2x+3$,then $NC = AC - AN = (2x+3) - (x+5) = x-2$.Substituting these into the ratio: $\frac{x+3}{x-3} = \frac{x+5}{x-2}$.Cross-multiplying gives: $(x+3)(x-2) = (x+5)(x-3)$.Expanding both sides: $x^2 - 2x + 3x - 6 = x^2 - 3x + 5x - 15$.Simplifying: $x^2 + x - 6 = x^2 + 2x - 15$.Subtracting $x^2$ from both sides: $x - 6 = 2x - 15$.Rearranging terms: $15 - 6 = 2x - x$.Therefore,$x = 9$.

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

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