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extra practice in exercises 1 and 2, complete the proof. 1. given $over…

Question

extra practice
in exercises 1 and 2, complete the proof.

  1. given $overline{ab}$ and $overline{cd}$ bisect each other at point $m$ and $overline{bm}=overline{cm}$.

prove $ab = am + dm$
statements

  1. $overline{bm}congoverline{cm}$
  2. $overline{cm}congoverline{dm}$
  3. $overline{bm}congoverline{dm}$
  4. $bm = dm$

5.

  1. $ab = am + dm$

reasons

  1. given

2.
3.
4.

  1. segment addition postulate (post. 1.2)

6.

Explanation:

Step1: Definition of bisecting segments

Since $\overline{AB}$ and $\overline{CD}$ bisect each other at point $M$, when a segment is bisected, the two resulting segments are congruent. So, $\overline{CM}\cong\overline{DM}$ because $M$ is the mid - point of $\overline{CD}$.

Step2: Transitive property of congruence

We know that $\overline{BM}\cong\overline{CM}$ (given) and $\overline{CM}\cong\overline{DM}$, so by the transitive property of congruence, $\overline{BM}\cong\overline{DM}$.

Step3: Definition of congruent segments

If two segments are congruent, their lengths are equal. So, if $\overline{BM}\cong\overline{DM}$, then $BM = DM$.

Step4: Segment addition postulate

By the segment addition postulate, $AB=AM + BM$.

Step5: Substitution

Since $BM = DM$, we substitute $BM$ with $DM$ in the equation $AB=AM + BM$. So, $AB=AM + DM$.

Answer:

STATEMENTSREASONS
2. $\overline{CM}\cong\overline{DM}$2. Definition of bisecting segments (since $M$ bisects $\overline{CD}$)
3. $\overline{BM}\cong\overline{DM}$3. Transitive property of congruence
4. $BM = DM$4. Definition of congruent segments
5. $AB=AM + BM$5. Segment Addition Postulate (Post. 1.2)
6. $AB=AM + DM$6. Substitution ($BM = DM$)