Question

1. proof $overline{ac} cong overline{gi}$, $overline{fe} cong overline{lk}$, and $ac + cf + fe = gi + il + lk$

prove that $overline{cf} cong overline{il}$.

1. proof consider $overline{ps}$.

a. complete the two - column proof.
given: $overline{pq} cong overline{rs}$
prove: $overline{pr} cong overline{qs}$

statement	reason
2. ?	2. definition of congruent segments
3. $pq + qr = pr$ and $qr + rs = qs$	3. ?
4. $rs + qr = pr$	4. ?
5. $qr + rs = pr$	5. ?
6. $qs = pr$	6. ?
7. $pr = qs$	7. symmetric property of equality
8. ?	8. definition of congruent segments

b. can it also be proved that $overline{pq} cong overline{rs}$ if $overline{pr} cong overline{qs}$? write a narrative proof to explain your reasoning.

Explanation:

Part 10a: Two - Column Proof

Step 1: Identify the given

The first statement is given as $\overline{PQ}\cong\overline{RS}$, so the reason for statement 1 is "Given".

Step 2: Use the definition of congruent segments

By the definition of congruent segments, if $\overline{PQ}\cong\overline{RS}$, then $PQ = RS$. So the statement for step 2 is $PQ=RS$.

Step 3: Identify the postulate for segment addition

The equations $PQ + QR=PR$ and $QR + RS = QS$ come from the Segment Addition Postulate, which states that if a point $Q$ is between $P$ and $R$, then $PQ+QR = PR$ (and similarly for $Q$, $R$, $S$). So the reason for step 3 is "Segment Addition Postulate".

Step 4: Substitute $PQ$ with $RS$

Since we know from step 2 that $PQ = RS$, we can substitute $PQ$ with $RS$ in the equation $PQ+QR = PR$. So we get $RS + QR=PR$. The reason for this step is "Substitution Property of Equality".

Step 5: Use the Commutative Property of Addition

The Commutative Property of Addition states that $a + b=b + a$. So $RS+QR = QR + RS$. So the reason for step 5 is "Commutative Property of Addition".

Step 6: Substitute and use transitive property

From step 3, we have $QR + RS=QS$, and from step 5 we have $QR + RS = PR$. By the Transitive Property of Equality (if $a = b$ and $b=c$, then $a = c$), we get $QS=PR$. The reason for step 6 is "Transitive Property of Equality".

Step 7: Symmetric Property of Equality

The Symmetric Property of Equality states that if $a = b$, then $b=a$. So from $QS = PR$, we get $PR = QS$.

Step 8: Use the definition of congruent segments

If $PR = QS$, then by the definition of congruent segments, $\overline{PR}\cong\overline{QS}$. So the statement for step 8 is $\overline{PR}\cong\overline{QS}$.

Filling in the two - column proof:

Statement	Reason
2. $PQ = RS$	2. Definition of congruent segments
3. $PQ + QR=PR$ and $QR + RS = QS$	3. Segment Addition Postulate
4. $RS + QR=PR$	4. Substitution Property of Equality
5. $QR + RS=PR$	5. Commutative Property of Addition
6. $QS = PR$	6. Transitive Property of Equality
7. $PR = QS$	7. Symmetric Property of Equality
8. $\overline{PR}\cong\overline{QS}$	8. Definition of congruent segments

Part 10b: Narrative Proof

To prove that $\overline{PQ}\cong\overline{RS}$ given that $\overline{PR}\cong\overline{QS}$:

1. Since $\overline{PR}\cong\overline{QS}$, by the definition of congruent segments, $PR = QS$.
2. By the Segment Addition Postulate, we know that $PR=PQ + QR$ and $QS=QR + RS$.
3. Substitute $PR$ and $QS$ in the equation $PR = QS$: $PQ+QR=QR + RS$.
4. Subtract $QR$ from both sides of the equation $PQ + QR=QR + RS$. By the Subtraction Property of Equality (if $a + b=c + b$, then $a=c$), we get $PQ=RS$.
5. Since $PQ = RS$, by the definition of congruent segments, $\overline{PQ}\cong\overline{RS}$.

Answer:

(10a two - column proof filled):

Statement	Reason
2. $PQ = RS$	2. Definition of congruent segments
3. $PQ + QR=PR$ and $QR + RS = QS$	3. Segment Addition Postulate
4. $RS + QR=PR$	4. Substitution Property of Equality
5. $QR + RS=PR$	5. Commutative Property of Addition
6. $QS=PR$	6. Transitive Property of Equality
7. $PR = QS$	7. Symmetric Property of Equality
8. $\overline{PR}\cong\overline{QS}$	8. Definition of congruent segments