Question

1. given: pq = pr prove: q is the mid - point of rh

statements reasons

1. pq = pr 1. given
2. 2. segment addition postulate
3. pq=pq + qr 3.
4. pq = qr 4.
5. 5. definition of midpoint
6. given: ab = cd, bd = de prove: ad = ce

statements reasons

1. ab = cd, bd = de 1. given
2. 2. definition of congruence
3. ab + bd = ad 3.
4. cd + de = ad 4.
5. 5. segment addition postulate
6. ad = ce 6.
7. definition of congruence
8. given: gi = jl, gh = kl prove: hi = jk

statements reasons

1. given

Explanation:

Step 1: Fill in statement 2 for first proof

By segment - addition postulate, if we consider the whole segment $PR$ composed of two sub - segments $PQ$ and $QR$, we have $PR=PQ + QR$. So the statement 2 is $PR=PQ + QR$.

Step 2: Fill in reason 3 for first proof

Since $PQ=PR$ (given) and $PR = PQ+QR$, by substitution property of equality, we can substitute $PR$ with $PQ$ in the equation $PR=PQ + QR$. So the reason 3 is "Substitution Property of Equality".

Step 3: Fill in reason 4 for first proof

If $PQ=PR$ and $PR=PQ + QR$, then subtracting $PQ$ from both sides of the equation $PR=PQ + QR$ (using the subtraction property of equality), we get $PQ=QR$. So the reason 4 is "Subtraction Property of Equality".

Step 4: Fill in statement 5 for first proof

The statement 5 is "$Q$ is the mid - point of $PR$" which follows from the previous step where $PQ = QR$ and the definition of mid - point.

Step 5: Fill in statement 2 for second proof

By the definition of congruence, if $AB\cong CD$ and $BD\cong DE$, then $AB = CD$ and $BD=DE$. So the statement 2 is $AB = CD, BD=DE$.

Step 6: Fill in reason 3 for second proof

The reason 3 for $AB + BD=AD$ is the segment - addition postulate which states that if $B$ is between $A$ and $D$, then $AB + BD=AD$.

Step 7: Fill in reason 4 for second proof

Since $AB = CD$ and $BD=DE$ (from step 2) and $AB + BD=AD$, by substitution property of equality, we can substitute $AB$ with $CD$ and $BD$ with $DE$ in the equation $AB + BD=AD$ to get $CD + DE=AD$. So the reason 4 is "Substitution Property of Equality".

Step 8: Fill in statement 5 for second proof

By the segment - addition postulate, $CD+DE = CE$.

Step 9: Fill in reason 6 for second proof

Since $AD=CD + DE$ and $CE=CD + DE$ (from previous steps), by the transitive property of equality, $AD = CE$. So the reason 6 is "Transitive Property of Equality".

Step 10: Fill in statement 7 for second proof

The statement 7 is $AD\cong CE$ which follows from $AD = CE$ and the definition of congruence.

Step 11: Fill in statements and reasons for third proof

Statements:

1. $\overline{GI}\cong\overline{JL},\overline{GH}\cong\overline{KL}$
2. $GI = JL, GH=KL$ (Definition of Congruence)
3. $GI=GH + HI$ (Segment - addition postulate)
4. $JL=JK + KL$ (Segment - addition postulate)
5. $GH + HI=JK + KL$ (Substitution Property of Equality, since $GI = JL$)
6. $HI=JK$ (Subtraction Property of Equality, subtracting $GH$ (which is equal to $KL$) from both sides)
7. $\overline{HI}\cong\overline{JK}$ (Definition of Congruence)

Reasons:

1. Given
2. Definition of Congruence
3. Segment - addition postulate
4. Segment - addition postulate
5. Substitution Property of Equality
6. Subtraction Property of Equality
7. Definition of Congruence

Answer:

First proof:

Statements	Reasons
2. $PR=PQ + QR$	2. Segment Addition Postulate
3. $PQ=PQ + QR$	3. Substitution Property of Equality
4. $PQ=QR$	4. Subtraction Property of Equality
5. $Q$ is the mid - point of $PR$	5. Definition of Midpoint

Second proof:

Statements	Reasons
2. $AB = CD, BD=DE$	2. Definition of Congruence
3. $AB + BD=AD$	3. Segment Addition Postulate
4. $CD + DE=AD$	4. Substitution Property of Equality
5. $CD+DE = CE$	5. Segment Addition Postulate
6. $AD = CE$	6. Transitive Property of Equality
7. $AD\cong CE$	7. Definition of Congruence

Third proof:

Statements	Reasons
2. $GI = JL, GH=KL$	2. Definition of Congruence
3. $GI=GH + HI$	3. Segment - addition postulate
4. $JL=JK + KL$	4. Segment - addition postulate
5. $GH + HI=JK + KL$	5. Substitution Property of Equality
6. $HI=JK$	6. Subtraction Property of Equality
7. $\overline{HI}\cong\overline{JK}$	7. Definition of Congruence