Question

given: $overline{lk}congoverline{np}$; m is the mid - point of $overline{kp}$; $overline{lm}congoverline{om}$; $overline{mn}congoverline{om}$
prove: $\triangle lkmcong\triangle npm$

1. $overline{lk}congoverline{np}$, m is the mid - point of $overline{kp}$. 1. given
2. definition of a mid - point
3. $overline{lm}congoverline{om}$, $overline{mn}congoverline{om}$

4.

1. $\triangle lkmcong\triangle npm$

Explanation:

Step1: State given information

Given $\overline{LK}\cong\overline{NP}$, M is mid - point of $\overline{KP}$, $\overline{LM}\cong\overline{OM}$, $\overline{MN}\cong\overline{OM}$

Step2: Apply mid - point definition

$KM = MP$ as M is mid - point of $\overline{KP}$

Step3: Restate given

$\overline{LM}\cong\overline{OM}$, $\overline{MN}\cong\overline{OM}$

Step4: Use transitive property

$\overline{LM}=\overline{MN}$

Step5: Apply SSS criterion

$\triangle LKM\cong\triangle NPM$

Answer:

1. Given
2. $KM = MP$; Since M is the mid - point of $\overline{KP}$, by the definition of a mid - point, a mid - point divides a line segment into two equal parts.
3. Given
4. $\overline{LM}=\overline{MN}$; Because $\overline{LM}\cong\overline{OM}$ and $\overline{MN}\cong\overline{OM}$, by the transitive property of congruence (if $a = c$ and $b = c$, then $a = b$).
5. $\triangle LKM\cong\triangle NPM$; By the Side - Side - Side (SSS) congruence criterion since $\overline{LK}\cong\overline{NP}$ (given), $KM = MP$ (from step 2) and $\overline{LM}=\overline{MN}$ (from step 4).