Question

1. given: 2pq = pr

prove: q is the mid - point of pr
statements reasons

1. 2pq = pr 1. given
2. pq + qr = pr 2. segment addition postulate
3. 2pq = 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. 7. definition of congruence
8. given: gi ≅ jl, gh ≅ kl

prove: hi ≅ jk
statements reasons

Explanation:

Problem 6

Step1: Substitute segment addition

Since $PQ + QR=PR$ and $2PQ = PR$, we substitute $PR$ in the first - equation. So, from $2PQ=PR$ and $PQ + QR = PR$, we get $2PQ=PQ + QR$ (by substitution property of equality).

Step2: Solve for $PQ$ and $QR$

Subtract $PQ$ from both sides of the equation $2PQ=PQ + QR$. We have $2PQ−PQ=PQ + QR−PQ$, which simplifies to $PQ = QR$.

Step3: Prove mid - point

By the definition of mid - point (a point that divides a segment into two congruent segments), since $PQ = QR$, $Q$ is the mid - point of $\overline{PR}$.

Step1: Use congruence definition

If $\overline{AB}\cong\overline{CD}$ and $\overline{BD}\cong\overline{DE}$, by the definition of congruence, $AB = CD$ and $BD=DE$.

Step2: Segment addition for $AD$

By the segment addition postulate, $AB + BD=AD$.

Step3: Substitute for $AD$

Since $AB = CD$ and $BD = DE$, we substitute into the equation $AB + BD=AD$ to get $CD+DE = AD$.

Step4: Segment addition for $CE$

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

Step5: Transitive property

Since $AD=CD + DE$ and $CE=CD + DE$, by the transitive property of equality, $AD = CE$.

Step6: Use congruence definition

Since $AD = CE$, by the definition of congruence, $\overline{AD}\cong\overline{CE}$.

Step1: Use congruence definition

Since $\overline{GI}\cong\overline{JL}$ and $\overline{GH}\cong\overline{KL}$, by the definition of congruence, $GI=JL$ and $GH = KL$.

Step2: Segment addition for $GI$ and $JL$

We know that $GI=GH + HI$ and $JL=JK + KL$ (by the segment addition postulate).

Step3: Substitute

Substitute $GI = JL$, $GH = KL$ into $GI=GH + HI$ and $JL=JK + KL$. We get $GH + HI=JK + KL$.

Step4: Subtract equal segments

Subtract $GH$ (which is equal to $KL$) from both sides of the equation $GH + HI=JK + KL$. So, $GH + HI−GH=JK + KL−KL$, which simplifies to $HI = JK$.

Step5: Use congruence definition

Since $HI = JK$, by the definition of congruence, $\overline{HI}\cong\overline{JK}$.

Answer:

1. Substitution Property of Equality
2. Subtraction Property of Equality
3. $Q$ is the mid - point of $\overline{PR}$

Problem 7