In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 3x - 19$,$MB = x - 3$,$MC = x - 5$,and $MD = 3$,find the value of $x$.
(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion.
Therefore,the ratios of corresponding sides are equal: $\frac{MA}{MC} = \frac{MB}{MD}$.
Substituting the given values: $\frac{3x - 19}{x - 5} = \frac{x - 3}{3}$.
Cross-multiplying gives: $3(3x - 19) = (x - 5)(x - 3)$.
$9x - 57 = x^2 - 3x - 5x + 15$.
$9x - 57 = x^2 - 8x + 15$.
Rearranging into a quadratic equation: $x^2 - 17x + 72 = 0$.
Factoring the quadratic: $(x - 8)(x - 9) = 0$.
Thus,$x = 8$ or $x = 9$.
Therefore,the ratios of corresponding sides are equal: $\frac{MA}{MC} = \frac{MB}{MD}$.
Substituting the given values: $\frac{3x - 19}{x - 5} = \frac{x - 3}{3}$.
Cross-multiplying gives: $3(3x - 19) = (x - 5)(x - 3)$.
$9x - 57 = x^2 - 3x - 5x + 15$.
$9x - 57 = x^2 - 8x + 15$.
Rearranging into a quadratic equation: $x^2 - 17x + 72 = 0$.
Factoring the quadratic: $(x - 8)(x - 9) = 0$.
Thus,$x = 8$ or $x = 9$.
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