Question

2 construct \\(\overline{be}\\) perpendicular to \\(\overleftrightarrow{cd}\\) such that point e is on \\(\overleftrightarrow{cd}).
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 measure (90^{circ}) (2, 3).
5 (cf = be) any point on one parallel line is the same distance from the other line on a perpendicular transversal (1, 2, 3).
6 (bc = bc) they are measures of the same segment.
7 (\triangle bcf\cong\triangle cbe)
8 (\angle fbc\cong\angle ecb) congruence (4, 6, 5)
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proofs with transformations
8:45

Explanation:

Step1: Identify given right - angles

We know that $\angle CFB = \angle BEC=90^{\circ}$ from steps 2 - 4.

Step2: Identify equal sides

$CF = BE$ (distance between parallel lines) and $BC = BC$ (common side).

Step3: Determine congruence criterion

We have two right - angled triangles $\triangle BCF$ and $\triangle CBE$. We have a right - angle in each triangle, a pair of equal non - hypotenuse sides ($CF = BE$) and a common side ($BC$). The Hypotenuse - Leg (HL) congruence criterion is used for right - angled triangles. Here, $BC$ is the hypotenuse for both right - angled triangles $\triangle BCF$ and $\triangle CBE$, and $CF$ and $BE$ are the legs.

Answer:

Hypotenuse - Leg