Question

given: \\(\overline{km} \cong \overline{nm}\\), \\(\overline{lk} \cong \overline{ln}\\)
prove: \\(\triangle lkm \cong \triangle lnm\\)

statements	reasons
2. \\(\overline{lk} \cong \overline{ln}\\)	2. given
3. \\(\overline{lm} \cong \overline{lm}\\)	3. reflexive
4. \\(\triangle lkm \cong \triangle lnm\\)	4. sss \\(\cong\\)

11 given: \\(\overline{ad} \cong \overline{ab}\\), \\(\overline{ac}\\) bisects \\(\angle dab\\)
prove: \\(\triangle adc \cong \triangle abc\\)

statements	reasons
2.	2.
3.	3.
4.	4.
5.	5.

12 given: \\(\overline{jm} \cong \overline{mn}\\), l is the midpoint of \\(\overline{jn}\\)
prove: \\(\triangle jlm \cong \triangle nlm\\)

statements	reasons
2. l is midpoint of \\(\overline{jn}\\)	2.
3.	3.
4.	4.
5.	5.

Explanation:

Step1: Analyze Given Information

We are given \( \overline{AD} \cong \overline{AB} \) and \( \overline{AC} \) bisects \( \angle DAB \). We need to prove \( \triangle ADC \cong \triangle ABC \).

Step2: State Angle Bisector Definition

Since \( \overline{AC} \) bisects \( \angle DAB \), by the definition of an angle bisector, \( \angle DAC \cong \angle BAC \).

Step3: Identify Common Side

The side \( \overline{AC} \) is common to both \( \triangle ADC \) and \( \triangle ABC \), so \( \overline{AC} \cong \overline{AC} \) (reflexive property of congruence).

Step4: Apply SAS Congruence Criterion

We have \( \overline{AD} \cong \overline{AB} \) (given), \( \angle DAC \cong \angle BAC \) (from angle bisector), and \( \overline{AC} \cong \overline{AC} \) (common side). By the Side - Angle - Side (SAS) congruence criterion, \( \triangle ADC \cong \triangle ABC \).

Now, filling in the two - column proof:

Statements	Reasons
2. \( \overline{AC} \) bisects \( \angle DAB \)	2. Given
3. \( \angle DAC \cong \angle BAC \)	3. Definition of angle bisector
4. \( \overline{AC} \cong \overline{AC} \)	4. Reflexive property of congruence
5. \( \triangle ADC \cong \triangle ABC \)	5. SAS congruence criterion

Answer:

The completed two - column proof is as shown above. For the blanks in the original problem:

1. \( \overline{AC} \) bisects \( \angle DAB \); Reason: Given

1. \( \angle DAC \cong \angle BAC \); Reason: Definition of angle bisector

1. \( \overline{AC} \cong \overline{AC} \); Reason: Reflexive property of congruence

1. \( \triangle ADC \cong \triangle ABC \); Reason: SAS congruence criterion