Question

given: ac ⊥ 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 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. ∠fcd = ∠fcg + ∠gce + ∠ecd = 180° by the definition of a straight angle, and ∠gce = 90° by definition of perpendicular lines. so, by the substitution property of equality, 180° = ∠fcg + 90° + ∠ecd. therefore, 90° - ∠fcg = ∠ecd by the subtraction property of equality. we also know that 180° = ∠fcg + 90° + ∠cge by the triangle - sum theorem, and by the substitution property of equality, ∠ecd = ∠cge by the definition of congruent angles. then, ∠ecd ≅ ∠cge by the definition of congruent angles. so, by aa, △gfc ∼ △cde. since gf • de = cd • fc by cross - product. finally, by the division property of equality, \\(\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: Recall slope - perpendicular lines relation

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of line \(AC\) be \(m_1=\frac{GF}{FC}\) and the slope of line \(BC\) be \(m_2 =-\frac{DE}{CD}\).

Step2: Use triangle - congruence and angle - relation

Since \(\triangle GFC\sim\triangle CDE\) (by AA similarity as angles are congruent). We know that \(\frac{GF}{FC}=\frac{CD}{DE}\).

Step3: Calculate the product of slopes

The product of the slopes \(m_1\times m_2=\frac{GF}{FC}\times(-\frac{DE}{CD})\). Substituting \(\frac{GF}{FC}=\frac{CD}{DE}\) into the product, we get \(\frac{CD}{DE}\times(-\frac{DE}{CD})=- 1\).

Answer:

The product of the slopes of lines \(AC\) and \(BC\) is - 1, which proves that \(AC\perp BC\).