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).
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 (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 (\boldsymbol{downarrow}). (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 (\boldsymbol{downarrow}). 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 ecdcongangle cgf) by the definition of congruent angles. (angle gfccongangle cde) because all right angles are congruent. so, by aa, (\triangle gfcsim\triangle cde). since (\boldsymbol{downarrow}), then (\frac{gf}{cb}=\frac{fc}{de}), or (gfcdot de = cdcdot 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 (\boldsymbol{downarrow}) 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 of lines (ac) and (bc) is (-1).

Explanation:

Step1: Analyze angle sum for straight line

A straight angle is \(180^\circ\). For \(\angle FCD\), since it's a straight angle (along the line with \(\angle FCG\) and \(\angle GCE\) and \(\angle ECD\)? Wait, no, the straight angle here: \(\angle FCD = 180^\circ\) by straight angle definition, and it's composed of \(\angle FCG+\angle GCE+\angle ECD\)? Wait, no, the text says "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 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 **subtraction property of equality** (subtract \(\angle FCG\) and \(90^\circ\) from both sides? Wait, rearranging \(180 = \angle FCG + 90 + \angle ECD\) gives \(\angle ECD=180 - 90 - \angle FCG = 90 - \angle FCG\), so subtraction property of equality.

Step2: Analyze triangle similarity and congruence

We know that \(\angle GFC\) and \(\angle CDE\) are right angles (since \(FG\) is horizontal, \(DE\) is vertical? Wait, \(FG\) is horizontal, \(FC\) is vertical? Wait, the diagram has right angles at \(F\) and \(D\). So \(\angle GFC = \angle CDE = 90^\circ\). We found \(\angle ECD = 90^\circ - \angle FCG\), and \(\angle CGF = 90^\circ - \angle FCG\) (since \(\angle GFC = 90^\circ\), so \(\angle CGF = 90^\circ - \angle FCG\) by triangle angle sum in \(\triangle FCG\)). So \(\angle ECD=\angle CGF\) (by substitution, both equal to \(90 - \angle FCG\)). Then, in \(\triangle GFC\) and \(\triangle CDE\), \(\angle GFC=\angle CDE = 90^\circ\) and \(\angle CGF=\angle ECD\), so by AA (Angle - Angle) similarity, \(\triangle GFC \sim \triangle CDE\).

Step3: Use similarity to get proportion

For similar triangles \(\triangle GFC \sim \triangle CDE\), the corresponding sides are proportional. So \(\frac{GF}{CD}=\frac{FC}{DE}\) (since in similar triangles, \(\frac{\text{corresponding side 1}}{\text{corresponding side 2}}=\frac{\text{corresponding side 3}}{\text{corresponding side 4}}\)). Cross - multiplying gives \(GF\cdot DE = CD\cdot FC\).

Step4: Relate slopes and product

The slope of \(AC\) (or \(GC\)) is \(\frac{GF}{FC}\), and the slope of \(BC\) (or \(CE\)) is \(\frac{-DE}{CD}\). The product of the slopes is \(\frac{GF}{FC}\cdot\frac{-DE}{CD}\). From \(GF\cdot DE = CD\cdot FC\), we can rearrange to \(\frac{GF}{FC}=\frac{CD}{DE}\). Substituting into the slope product: \(\frac{CD}{DE}\cdot\frac{-DE}{CD}=-1\). So the key steps: angle subtraction, AA similarity, proportion from similarity, and then slope product.

But the drop - down menus: Let's assume the first drop - down (after "by the") for the angle part is "subtraction property of equality", the second (for triangle congruence/similarity) when we have \(\angle ECD\cong\angle CGF\) (wait, \(\angle ECD = \angle CGF\) as we saw, and \(\angle GFC=\angle CDE = 90^\circ\), so by AA, \(\triangle GFC\sim\triangle CDE\). Then, since \(\angle GEC\) and \(\angle CDE\) are right angles? Wait, no, the text says "all right angles are congruent. So, by AA, \(\triangle GFC\sim\triangle CDE\). Since \(GF\cdot DE = CD\cdot FC\) 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 \(BC\) (which is \(\frac{-DE}{CD}\)) to get \(\frac{GF}{FC}\cdot\frac{-DE}{CD}=\frac{CD}{DE}\cdot\frac{-DE}{CD}=-1\).

So for the first blank (after "by the") in the angle part: subtraction property of equality. For the congruence of \(\angle ECD\) and \(\angle CGF\): they are equal (congruent) by substitution (since both equal \(90 - \angle FCG\)). For the triangle congruence/similarity: \(\triangle GFC\sim\triangle CDE\) by AA (two angles: right angle and the equal acute angle). Then, the proportion \(GF\cdot DE = CD\cdot FC\) is cross - product from similar triangles (\(\frac{GF}{CD}=\frac{FC}{DE}\) implies \(GF\cdot DE = CD\cdot FC\)). Then, multiplying the slopes: \(\frac{GF}{FC}\cdot\frac{-DE}{CD}=\frac{CD}{DE}\cdot\frac{-DE}{CD}=-1\) (by substituting \(\frac{GF}{FC}=\frac{CD}{DE}\) from the proportion).

Answer:

(Assuming the drop - down options: For the angle step, "subtraction property of equality"; for the angle congruence, "\(\angle ECD\cong\angle CGF\) by substitution"; for the triangle similarity, "\(\triangle GFC\sim\triangle CDE\) by AA"; for the proportion, "cross - product property"; and for the final step, "substitution" to get \(\frac{GF}{FC}\cdot\frac{-DE}{CD}=\frac{CD}{DE}\cdot\frac{-DE}{CD}\). But since the problem is to complete the proof, the key steps lead to the product of slopes being \(-1\) as shown. The final answer for the proof conclusion is that the product of the slopes of \(AC\) and \(BC\) is \(-1\), so the answer is \(\boldsymbol{-1}\) for the product, and the steps use properties like subtraction property of equality, AA similarity, cross - product, and substitution. But if we are to fill the blanks:

1. For \(\angle FCD = 180^\circ\) by straight angle, then \(90^\circ-\angle FCG=\angle ECD\) by subtraction property of equality.

1. \(\angle ECD\cong\angle CGF\) by substitution (since both equal \(90^\circ - \angle FCG\)).

1. \(\triangle GFC\sim\triangle CDE\) by AA (Angle - Angle) similarity (right angle and equal acute angle).

1. \(GF\cdot DE = CD\cdot FC\) by cross - product property (from \(\frac{GF}{CD}=\frac{FC}{DE}\)).

1. Then \(\frac{GF}{FC}\cdot\frac{-DE}{CD}=\frac{CD}{DE}\cdot\frac{-DE}{CD}=-1\) by substitution (substituting \(\frac{GF}{FC}=\frac{CD}{DE}\)) and simplification.

But the main goal is to prove the product of slopes is \(-1\), so the final answer for the product is \(-1\).)

\(\boldsymbol{-1}\)