Question

prove: the product of the slopes of lines ac and bc is -1. construct a horizontal line that passes through the intersection of lines ac and bc at point c. add a vertical line segment connecting the horizontal line to line ac called fg and another vertical line connecting the horizontal line to line bc called de. complete the proof. the slope of line ac or gc is \\(\frac{gf}{fc}\\) by definition of slope. the slope of line bc or ce is \\(\frac{-de}{cd}\\) by definition of slope. \\(\angle fcd = \angle fcg + \angle gce + \angle ecd\\) by dropdown options: definition of a straight angle, definition of congruent angles, angle addition, definition of perpendicular lines. \\(\angle gce = 90^\circ\\) by definition of perpendicular lines. therefore, \\(90^\circ - \angle fcg = \angle ecd\\) by the dropdown options: definition of a straight angle, definition of congruent angles, angle addition, definition of perpendicular lines. the triangle sum theorem, and by the subtraction substitution property of equality. then, \\(\angle ecd\\) are congruent. so, by aa, \\(\triangle gfc \sim \triangle cdi\\) note: likely typo, maybe \\(\triangle cde\\) gf • de = cd • fc by cross product. finally, by the division property of equality, \\(\frac{gf}{fc} = \frac{cd}{de}\\). we can multiply both sides by the slope of line bc using the dropdown: substitution property of equality to get \\(\frac{gf}{fc} \bullet \frac{-de}{cd} = \frac{cd}{de} \bullet \frac{-de}{cd}\\). simplify so that \\(\frac{gf}{fc} \bullet \frac{-de}{cd} = -1\\). this shows that the product of the slopes of lines ac and bc is -1.

Explanation:

Step1: Analyze Angle Sum/Straight Angle

A straight angle is \(180^\circ\), so \(\angle FCD = 180^\circ\) (definition of a straight angle). Given \(\angle FCD=\angle FCG + \angle GCE+\angle ECD\), and \(\angle GCE = 90^\circ\) (perpendicular lines), we use the angle addition postulate here. So the first blank (for \(\angle FCD = \angle FCG+\angle GCE+\angle ECD\)) is filled by "angle addition postulate".

Step2: Subtract to Find Congruent Angles

We know \(180^\circ=\angle FCG + 90^\circ+\angle ECD\) (from straight angle and \(\angle GCE = 90^\circ\)). Subtract \(90^\circ\) from both sides: \(90^\circ-\angle FCG=\angle ECD\) (subtraction property of equality). Also, since \(\angle FCG\) and \(\angle CGF\) are related? Wait, no—wait, \(\angle GCE = 90^\circ\) (perpendicular), so \(\angle FCG+\angle GCE = 90^\circ\)? Wait, no, the key is that \(\angle FCG\) and \(\angle ECD\) are related by subtraction. Then, \(\angle ECD=\angle CGF\) (by...? Wait, the triangles: \(\triangle GFC\) and \(\triangle CD E\) – \(\angle GFC=\angle CDE = 90^\circ\) (perpendicular lines), and \(\angle FCG=\angle ECD\) (from angle subtraction). So by AA similarity, \(\triangle GFC\sim\triangle CDE\).

Step3: Slope Relationship

Slope of \(GC\) (part of \(AC\)) is \(\frac{GF}{FC}\) (rise over run), slope of \(CE\) (part of \(BC\)) is \(\frac{-DE}{CD}\) (negative because direction). Since \(\triangle GFC\sim\triangle CDE\), \(\frac{GF}{FC}=\frac{CD}{DE}\) (corresponding sides of similar triangles). Then, multiply slopes: \(\frac{GF}{FC}\times\frac{-DE}{CD}=\frac{CD}{DE}\times\frac{-DE}{CD}=-1\).

Answer:

The key steps involve using the angle addition postulate for the angle sum, subtraction property of equality to find congruent angles, AA similarity for triangles, and then multiplying slopes of similar triangle sides to get \(-1\). The main concepts are angle addition, triangle similarity, and slope definition. The final proof shows the product of slopes is \(-1\), so the answer (for the proof completion) follows these steps, with the angle addition postulate, subtraction property, AA similarity, and slope multiplication leading to \(-1\).

(Note: If the question was to fill the blanks, the first blank for \(\angle FCD=\angle FCG+\angle GCE+\angle ECD\) is "angle addition postulate", the next for \(90^\circ - \angle FCG=\angle ECD\) is "subtraction property of equality", then triangle similarity (AA) for \(\triangle GFC\sim\triangle CDE\), and then slope multiplication gives \(-1\).)

But since the problem is to prove the product of slopes is \(-1\), the final answer (the conclusion) is that the product of the slopes of \(AC\) and \(BC\) is \(-1\), which is shown through similar triangles and slope definition.

Final Answer (for the proof's conclusion):

The product of the slopes of lines \(AC\) and \(BC\) is \(\boldsymbol{-1}\).