Given segments of length $$ \mathrm{a}, \mathrm{b} $$, and c as shown, construct a segment of length x so that $$ \frac{a}{\mathrm{~b}}=\frac{\mathrm{x}}{\mathrm{c}} $$. Let b and a be end-to-end on a side of an arbitrary angle. Call the vertex of the angle A , the point between b and $$ \mathrm{a} B $$, and the end of a C . Describe how to construct a segment of length x that satisfies the given proportion using this angle. A. On the other side of the angle, mark D such that $$ \mathrm{DA}=\mathrm{c} $$. Connect B with D , and draw a line through C that is parallel to $$ \overrightarrow{\mathrm{BD}} $$, labeling its intersection B. On the other side of the angle, mark D such that $$ \mathrm{DA}=\mathrm{c} $$. Connect C with D . Then $$ \mathrm{CD}=\mathrm{x} $$. C. On the other side of the angle, mark D such that $$ \mathrm{DA}=\mathrm{c} $$. Connect B with D . Then $$ \mathrm{BD}=\mathrm{x} $$. D. On the other side of the angle, mark D such that $$ \mathrm{DA}=\mathrm{C} $$. Connect C with D , and draw a line through B that is perpendicular to $$ \overrightarrow{C D} $$, labeling its intersection with $$ \overrightarrow{A D} $$ as E . Then $$ \mathrm{DE}=\mathrm{x} $$. E. It is not possible to construct the described segment.

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UpStudy AI · Accepted Answer

1. Draw an arbitrary angle with vertex $$ A $$.  
2. On one side of the angle, mark points $$ B $$ and $$ C $$ with $$ AB = b $$ and $$ BC = a $$ (so that $$ B $$ lies between $$ A $$ and $$ C $$).  
3. On the other side of the angle starting at $$ A $$, mark point $$ D $$ such that $$ AD = c $$.  
4. Connect point $$ B $$ to point $$ D $$ with a straight line.  
5. Through point $$ C $$ (on the side where $$ BC = a $$), draw a line parallel to the segment $$ BD $$. Let this parallel line meet the ray $$ AD $$ at point $$ E $$.

By the intercept (or Thales') theorem, triangles $$ ABC $$ and $$ ADE $$ are similar, which gives the relation  
$$
\frac{a}{b} = \frac{AE}{AD}.
$$
Since $$ AD = c $$, if we define $$ AE = x $$, then we have  
$$
\frac{a}{b} = \frac{x}{c}.
$$
Thus, the segment $$ AE $$ (or the portion marked as $$ x $$) satisfies the desired proportion.

The correct construction is described in Option A.
To construct a segment of length $$ x $$ such that $$ \frac{a}{b} = \frac{x}{c} $$, follow these steps: 1. Draw an angle with vertex $$ A $$. 2. On one side, mark points $$ B $$ and $$ C $$ with $$ AB = b $$ and $$ BC = a $$. 3. On the other side, mark point $$ D $$ with $$ AD = c $$. 4. Draw a line from $$ B $$ to $$ D $$. 5. Draw a line through $$ C $$ parallel to $$ BD $$, intersecting $$ AD $$ at point $$ E $$. The segment $$ AE $$ (or $$ x $$) will satisfy the proportion $$ \frac{a}{b} = \frac{x}{c} $$. Therefore, the correct answer is Option A.

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