Question

select the correct answer from each drop-down menu.
given: \\(\overline{ac} \perp \overline{bc}\\)
prove: the product of the slopes of lines \\(ac\\) and \\(bc\\) is \\(-1\\).
image of geometric figure with lines and points
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 \\(\overline{fg}\\) and another vertical line connecting the horizontal line to line \\(bc\\) called \\(\overline{de}\\).
complete the proof.
the slope of line \\(ac\\) or \\(\overleftrightarrow{gc}\\) is \\(\frac{gf}{fc}\\) by definition of slope. the slope of line \\(bc\\) or \\(\overleftrightarrow{ce}\\) is \\(\frac{de}{cd}\\) by definition of slope.
\\(\angle fcd = \angle fcg + \angle gce + \angle bcd\\) by drop-down, \\(\angle fcd = 180^\circ\\) by the definition of a straight angle, and \\(\angle gce = 90^\circ\\) by definition of perpendicular lines. so, by the substitution property of equality, \\(180^\circ = \angle fcg + 90^\circ + \angle ecd\\). therefore, \\(90^\circ - \angle fcg = \angle ecd\\) by the drop-down. we also know that \\(180^\circ = \angle fcg + 90^\circ + \angle cgf\\) by the triangle sum theorem, and by the subtraction property of equality, \\(90^\circ - \angle fcg = \angle cgf\\). therefore, \\(\angle ecd = \angle cgf\\) by the substitution property of equality. then, \\(\angle ecd \cong \angle cgf\\) by the definition of congruent angles. \\(\angle qfc \cong \angle cde\\) because all right angles are congruent. so, by aa, \\(\triangle gfc \sim \triangle cde\\). since drop-down, then \\(\frac{gf}{cd} = \frac{fc}{de}\\), or \\(gf \cdot de = cd \cdot 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 drop-down to get \\(\frac{gf}{fc} \cdot \frac{de}{cd} = \frac{cd}{de} \cdot \frac{de}{cd}\\). simplify so that \\(\frac{gf}{fc} \cdot \frac{de}{cd} = -1\\). this shows that the product of the slopes...
drop-down options: substitution property of equality, transitive property of equality, addition property of equality, multiplication property of equality

Explanation:

Step1: Analyze the first drop - down (angle addition)

The equation \(\angle FCD=\angle FCG+\angle GCE+\angle BCD\) is formed by the angle addition postulate, which states that if a point lies in the interior of an angle, the sum of the adjacent angles is equal to the larger angle. So the first drop - down should be filled with the angle addition postulate.

Step2: Analyze the second drop - down (subtraction property)

We have \(180^{\circ}=\angle FCG + 90^{\circ}+\angle ECD\). To get \(90^{\circ}-\angle FCG=\angle ECD\), we subtract \(\angle FCG\) and \(90^{\circ}\) from both sides of the equation, which is the subtraction property of equality.

Step3: Analyze the third drop - down (similar triangles ratio)

Since \(\triangle GFC\sim\triangle CDE\) by AA similarity (two right angles and a pair of congruent angles), the ratios of corresponding sides are equal. So \(\frac{GF}{CD}=\frac{FC}{DE}\) (corresponding sides of similar triangles are proportional).

Step4: Analyze the fourth drop - down (multiplication property)

We want to multiply both sides of \(\frac{GF}{FC}=\frac{CD}{DE}\) by the slope of \(BC\) (which is \(\frac{DE}{CD}\)) to get \(\frac{GF}{FC}\cdot\frac{DE}{CD}=\frac{CD}{DE}\cdot\frac{DE}{CD}\). This is the multiplication property of equality, which allows us to multiply both sides of an equation by the same non - zero number.

Answer:

First drop - down: Angle Addition Postulate; Second drop - down: Subtraction Property of Equality; Third drop - down: Corresponding Sides of Similar Triangles are Proportional; Fourth drop - down: Multiplication Property of Equality