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Math  /  Geometry

QuestionFill in the missing statement and reason of the proof below. Given: FD‾≅FC‾\overline{F D} \cong \overline{F C} and ∠FDE≅∠FCE\angle F D E \cong \angle F C E. Prove: △ADB≅△BCA\triangle A D B \cong \triangle B C A. p Statement FD‾≅FC‾\overline{F D} \cong \overline{F C} ∠FDE≅∠FCE\angle F D E \cong \angle F C E ∠F≅∠F\angle F \cong \angle F △ACF≅△BDF\triangle A C F \cong \triangle B D F
Reason
Given
Reflexive Property ASA
Type of Statement ∠DAB≅∠CBAAB‾≅AB‾\begin{aligned} \angle D A B & \cong \angle C B A \\ \overline{A B} & \cong \overline{A B} \end{aligned} AD‾≅BC‾△ADB≅△BCA\begin{array}{c} \overline{A D} \cong \overline{B C} \\ \triangle A D B \cong \triangle B C A \end{array}
In a triangle, angles opposite of congruent sides are congruent
Reflexive Property Congruent segments subtracted from congruent segments form congruent segments
SAS

Studdy Solution

STEP 1

1. We are given that FD‾≅FC‾\overline{F D} \cong \overline{F C} and ∠FDE≅∠FCE\angle F D E \cong \angle F C E.
2. We need to prove △ADB≅△BCA\triangle A D B \cong \triangle B C A.

STEP 2

1. Identify the missing statement and reason.
2. Complete the proof using congruence postulates.

STEP 3

Identify the missing statement and reason:
The missing statement is: ∠DAB≅∠CBA\angle D A B \cong \angle C B A
The reason for this statement is: In a triangle, angles opposite of congruent sides are congruent.

STEP 4

Complete the proof using congruence postulates:
1. Given: FD‾≅FC‾\overline{F D} \cong \overline{F C}
2. Given: ∠FDE≅∠FCE\angle F D E \cong \angle F C E
3. ∠F≅∠F\angle F \cong \angle F by the Reflexive Property
4. △ACF≅△BDF\triangle A C F \cong \triangle B D F by ASA (Angle-Side-Angle)
5. ∠DAB≅∠CBA\angle D A B \cong \angle C B A because in a triangle, angles opposite of congruent sides are congruent
6. AB‾≅AB‾\overline{A B} \cong \overline{A B} by the Reflexive Property
7. AD‾≅BC‾\overline{A D} \cong \overline{B C} because congruent segments subtracted from congruent segments form congruent segments
8. △ADB≅△BCA\triangle A D B \cong \triangle B C A by SAS (Side-Angle-Side)

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