Question

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given that lines ( l ) and ( n ) are perpendicular, which of the following is part of the proof that ( l ) and ( n ) have slopes that are opposite reciprocals?
(1 point)
( angle abccongangle dae )
( \triangle abc ) is similar to ( \triangle eda )
( \frac{bc}{ca}=-\frac{de}{ad} )
the slope of line ( l ) is ( \frac{bc}{ca} ) and the slope of line ( n ) is ( \frac{de}{ad} )

Explanation:

Step1: Analyze the angles

Since \( \angle ABC=\angle EDA = 90^{\circ}\) (right angles) and \( \angle BAC+\angle CAD = 90^{\circ}\), \( \angle CAD+\angle ADE=90^{\circ}\), so \( \angle BAC=\angle ADE\) (by the property that if two angles are complementary to the same angle, they are equal).

Step2: Use the AA (Angle - Angle) similarity criterion

In \( \triangle ABC\) and \( \triangle EDA\), we have two pairs of equal angles (\( \angle ABC=\angle EDA\) and \( \angle BAC=\angle ADE\)). By the AA similarity criterion for triangles, \( \triangle ABC\sim\triangle EDA\).

Step3: Relate to slope formula

The slope of a line is given by \(m = \frac{\text{rise}}{\text{run}}\). For line \(l\) (assuming a coordinate - like interpretation from the geometric figure), if we consider \( \triangle ABC\), the slope \(m_{l}=\frac{BC}{AB}\) (in a coordinate - based slope calculation, \(AB\) is the run and \(BC\) is the rise). For line \(n\), considering \( \triangle EDA\), \(m_{n}=\frac{DE}{AD}\). Since \( \triangle ABC\sim\triangle EDA\), we have \(\frac{BC}{AB}=-\frac{AD}{DE}\) (the negative sign comes from the direction of the lines in a coordinate system, as one line has a positive slope - like ratio and the other has a negative one due to perpendicularity). If we assume \(AB = AD\) (by the geometric construction related to the perpendicular lines and the similar triangles), we can show that the slopes are negative reciprocals. But the key step in the proof is establishing the similarity of the two right - angled triangles \( \triangle ABC\) and \( \triangle EDA\)

• Option A: \( \angle ABC\cong\angle DAE\) is incorrect. \( \angle ABC = 90^{\circ}\) and \( \angle DAE\) is not a right angle.
• Option C: \(\frac{BC}{CA}=-\frac{DE}{AD}\) is not a direct part of the standard proof for slopes of perpendicular lines. The slope formula is based on the ratio of the legs of the right - angled triangles (not the hypotenuse and a leg).
• Option D: The slope of line \(l\) is \(\frac{BC}{AB}\) (not \(\frac{BC}{CA}\)) and the slope of line \(n\) is \(-\frac{AD}{DE}\) (not \(\frac{DE}{AD}\)) when using the correct slope formula \(m=\frac{\text{rise}}{\text{run}}\) for non - vertical lines in a coordinate - like geometric interpretation.

Answer:

B. \( \triangle ABC \) is similar to \( \triangle EDA \)