Chapter 4: Q9. (page 144)

In many proofs you may find that different methods can be used. You may not know in advance which method will be better. There are two possible pairs of overlapping triangles that could be used in this proof. To compare the two methods, write a two-column proof for each plan.
Given: $\bar{PR} ≅ \bar{PQ}$ ; $\bar{SR} ≅ \bar{TQ}$
Prove: $\bar{QS} ≅ \bar{RT}$
a. plan for proof: show that $Δ RQS ≅ Δ QRT$ by SAS.
b. plan for proof: show that $Δ PQS ≅ Δ PRT$ by SAS.

Expert verified

a. The two-column proof is:

Statements	Reasons
$\bar{P R} ≅ \bar{P Q}, \bar{S R} ≅ \bar{T Q}$	Given
$∠ P R Q ≅ ∠ P Q R$	angles opposite to the equal sides $P R$ and $P Q$ will also be equal
$Δ R Q S ≅ Δ Q R T$	SAS postulate

b. The two-column proof is:

Statements	Reasons
$\bar{P R} ≅ \bar{P Q}, \bar{S R} ≅ \bar{T Q}$	Given
$P R = P S + S R, P Q = P T + T Q$	Segment addition postulate
$P T ≅ P S$	As, $P R = P S + S R$ , $P Q = P T + T Q$ , $\bar{P R} ≅ \bar{P Q}$ and $\bar{S R} ≅ \bar{T Q}$
$Δ P Q S ≅ Δ P R T$	SAS postulate

Step by step solution

The given diagram is:

It is being given that $\bar{P R} ≅ \bar{P Q}$ and $\bar{S R} ≅ \bar{T Q}$ .

As, $\bar{P R} ≅ \bar{P Q}$ , therefore,the angles opposite to the equal sides will also be equal.

Therefore, $∠ P R Q ≅ ∠ P Q R$ .

In the triangles width="48" height="19" role="math">ΔRQSand $Δ Q R T$ , it can be noticed that:

$\bar{S R} ≅ \bar{T Q}, ∠ P R Q ≅ ∠ P Q R$ and $R Q ≅ R Q$ .

Therefore, the triangles $Δ R Q S$ and $Δ Q R T$ are the congruent triangles by using SAS postulate.

Therefore, $Δ R Q S ≅ Δ Q R T$ .

The two-column proof is:

Statements	Reasons
$\bar{P R} ≅ \bar{P Q}, \bar{S R} ≅ \bar{T Q}$	Given
$∠ P R Q ≅ ∠ P Q R$	angles opposite to the equal sides $P R$ and $P Q$ will also be equal
$Δ R Q S ≅ Δ Q R T$	SAS postulate

The given diagram is:

It is being given that $\bar{P R} ≅ \bar{P Q}$ and $\bar{S R} ≅ \bar{T Q}$ .

From the given diagram it can be noticed that:

$P R = P S + S R$ and $P Q = P T + T Q$

As, $\bar{P R} ≅ \bar{P Q}$ , therefore it can be obtained that:

$\begin{matrix} P R = P Q \\ P S + S R = P T + T Q \\ P S + T Q = P T + T Q \\ P S = P T \end{matrix}$

Therefore, $P T ≅ P S$ .

In the triangles $Δ P Q S$ and $Δ P R T$ , it can be noticed that:

$\bar{P R} ≅ \bar{P Q}$ , $∠ P ≅ ∠ P$ and $P T ≅ P S$ .

Therefore, the triangles $Δ P Q S$ and $Δ P R T$ are the congruent triangles by using SAS postulate.

Therefore, $Δ P Q S ≅ Δ P R T$ .

The two-column proof is:

Statements	Reasons
$\bar{P R} ≅ \bar{P Q}, \bar{S R} ≅ \bar{T Q}$	Given
$P R = P S + S R, P Q = P T + T Q$	Segment addition postulate
$P T ≅ P S$	As, $P R = P S + S R, P Q = P T + T Q, \bar{P R} ≅ \bar{P Q}$ and $\bar{S R} ≅ \bar{T Q}$
$Δ P Q S ≅ Δ P R T$	SAS postulate