Question

given: nl bisects ∠knm and ∠klm prove: △nkl≅△nml
statement: nl bisects ∠knm and ∠klm, ∠lnk≅∠mnl, ∠kln≅∠mln, nl≅nl, △nkl≅△nml
reason: given, def of angle bisector, reflexive property, asa

Explanation:

Step1: Use angle - bisector definition

Since $\overline{NL}$ bisects $\angle KNM$ and $\angle KLM$, by the definition of an angle - bisector, we have $\angle LNK\cong\angle LNM$ and $\angle KLN\cong\angle MLN$.

Step2: Apply reflexive property

The side $\overline{NL}\cong\overline{NL}$ by the reflexive property of congruence.

Step3: Use ASA (Angle - Side - Angle)

In $\triangle NKL$ and $\triangle NML$, we have $\angle KLN\cong\angle MLN$ (from angle - bisector), $\overline{NL}\cong\overline{NL}$ (reflexive property), and $\angle LNK\cong\angle LNM$ (from angle - bisector). So, by the ASA congruence criterion, $\triangle NKL\cong\triangle NML$.

Answer:

The proof is completed as shown above to prove $\triangle NKL\cong\triangle NML$ using the angle - bisector definition, reflexive property, and ASA congruence criterion.