Question

3 construct \\(\overline{cf}\\) perpendicular to \\(\overleftrightarrow{ab}\\) such that point f is on \\(\overleftrightarrow{ab}\\).
4 \\(m\angle cfb = m\angle bec = 90^{\circ}\\) all perpendicular angles meas 3).
5 \\(cf = \quad\\) any point on one parallel line is distance from the other line on perpendicular transversal (1, 2.
6 \\(bc = bc\\) they are measures of the same
7 \\(\triangle bcf \cong \triangle cbe\\) \\(\quad\\) congruence (4, 6, 5)
8 \\(\angle fbc \cong \angle ecb\\) corresponding parts of congruer are congruent (7).

Explanation:

Step-by-Step Explanation:

Step 1: Analyze Statement 5

The right - hand side explanation for statement 5 says "Any point on one parallel line is equidistant from the other line on the perpendicular transversal". Also, from the diagram (implied by the construction of perpendiculars), we know that \(CF\) and \(BE\) are both perpendicular to \(AB\) (and if \(AB\) and another line are parallel, but from the congruence later, we can infer that \(CF = BE\) because of the equidistant property of parallel lines. So for statement 5, \(CF=BE\).

Step 2: Analyze Statement 7

We have two right triangles \(\triangle BCF\) and \(\triangle CBE\). We know that \(\angle CFB=\angle BEC = 90^{\circ}\) (statement 4), \(BC = BC\) (statement 6, reflexive property), and \(CF = BE\) (statement 5). The congruence criterion that uses a right angle, a hypotenuse, and a leg is the Hypotenuse - Leg (HL) congruence criterion. So \(\triangle BCF\cong\triangle CBE\) by HL congruence.

Final Answers:

For statement 5: \(BE\)

For statement 7: HL (Hypotenuse - Leg)

Answer:

Step-by-Step Explanation:

Step 1: Analyze Statement 5

The right - hand side explanation for statement 5 says "Any point on one parallel line is equidistant from the other line on the perpendicular transversal". Also, from the diagram (implied by the construction of perpendiculars), we know that \(CF\) and \(BE\) are both perpendicular to \(AB\) (and if \(AB\) and another line are parallel, but from the congruence later, we can infer that \(CF = BE\) because of the equidistant property of parallel lines. So for statement 5, \(CF=BE\).

Step 2: Analyze Statement 7

We have two right triangles \(\triangle BCF\) and \(\triangle CBE\). We know that \(\angle CFB=\angle BEC = 90^{\circ}\) (statement 4), \(BC = BC\) (statement 6, reflexive property), and \(CF = BE\) (statement 5). The congruence criterion that uses a right angle, a hypotenuse, and a leg is the Hypotenuse - Leg (HL) congruence criterion. So \(\triangle BCF\cong\triangle CBE\) by HL congruence.

Final Answers:

For statement 5: \(BE\)

For statement 7: HL (Hypotenuse - Leg)