Question

andrea
line segment \\( \overline { c d } \\) is the perpendicular bisector of line segment \\( \overline { a b } \\). which statement shows why \\( \overline { a e } \cong \overline { b e } \\)?
(a) \\( m \angle e f a = m \angle e f b = 90 ^ { \circ } \\) by definition of perpendicular so \\( \angle e f a \cong \angle e f b \\)
\\( \overline { a f } \cong \overline { b f } \\) by the definition of midpoint
\\( \triangle e a f \cong \triangle e b f \\) by hypotenuse-leg
\\( \overline { a e } \cong \overline { b e } \\) because they are corresponding parts of congruent triangles
(b) \\( m \angle e f a + m \angle e f b = 180 ^ { \circ } \\) so \\( \triangle a e b \\) must be isosceles
\\( \overline { a e } \cong \overline { b e } \\) because the legs of isosceles triangles are congruent
(c) \\( m \angle e f a = m \angle e f b = 90 ^ { \circ } \\) by definition of perpendicular so \\( \angle e f a \cong \angle e f b \\)
\\( \overline { a f } \cong \overline { b f } \\) by the definition of midpoint
\\( \overline { f f } \cong \overline { f f } \\) by the reflexive property

Explanation:

Step1: Analyze Option A

• \(m\angle EFA = m\angle EFB=90^{\circ}\) (definition of perpendicular, so \(\angle EFA\cong\angle EFB\))
• \(\overline{AF}\cong\overline{BF}\) (definition of mid - point)
• In right - triangles \(\triangle EAF\) and \(\triangle EBF\), we can use the Hypotenuse - Leg (HL) congruence criterion. If we consider \(EF\) as a common leg and \(AE\) and \(BE\) as hypotenuses. Since \(\triangle EAF\cong\triangle EBF\) (by HL), then \(\overline{AE}\cong\overline{BE}\) (corresponding parts of congruent triangles).

Step2: Analyze Option B

• \(m\angle EFA + m\angle EFB = 180^{\circ}\) only tells us that \(\angle EFA\) and \(\angle EFB\) are supplementary. Just because two angles are supplementary does not imply that \(\triangle AEB\) is isosceles. The logic \(m\angle EFA + m\angle EFB = 180^{\circ}\Rightarrow\triangle AEB\) is isosceles is incorrect.

Step3: Analyze Option C

• While \(m\angle EFA = m\angle EFB = 90^{\circ}\) (so \(\angle EFA\cong\angle EFB\)), \(\overline{AF}\cong\overline{BF}\) (mid - point), and \(\overline{EF}\cong\overline{EF}\) (reflexive property) gives us \(\triangle EAF\cong\triangle EBF\) by Side - Angle - Side (SAS) congruence criterion. But the option does not complete the reasoning by stating that \(\overline{AE}\cong\overline{BE}\) as corresponding parts of congruent triangles.

Answer:

A. \(m\angle EFA=m\angle EFB = 90^{\circ}\) by definition of perpendicular so \(\angle EFA\cong\angle EFB\), \(\overline{AF}\cong\overline{BF}\) by the definition of midpoint, \(\triangle EAF\cong\triangle EBF\) by Hypotenuse - Leg, \(\overline{AE}\cong\overline{BE}\) because they are corresponding parts of congruent triangles.