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In respects to the characteristics of the diagram, the following statement holds true.
PQ≅SRandQR≅PS
Two proofs will be provided for this theorem.
This theorem can be proven by placing the parallelogram on a coordinate plane. For simplicity, vertex P will be placed at the origin and vertex S on the x-axis.
m=x2−x1y2−y1 | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
PQ | P(0,0) and Q(b,c) | mPQ=b−0c−0 | mPQ=bc |
SR | S(a,0) and R(x,c) | mSR=x−ac−0 | mSR=x−ac |
LHS⋅b=RHS⋅b
ca⋅b=ca⋅b
LHS⋅(x−a)=RHS⋅(x−a)
LHS/c=RHS/c
LHS+a=RHS+a
Commutative Property of Addition
Substitute values
Subtract terms
d=(x2−x1)2+(y2−y1)2 | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
PQ | P(0,0) and Q(b,c) | PQ= (b−0)2+(c−0)2 | PQ=b2+c2 |
QR | Q(b,c) and R(a+b,c) | QR= (a+b−b)2+(c−c)2 | QR=a |
SR | S(a,0) and R(a+b,c) | SR= (a+b−a)2+(c−0)2 | SR=b2+c2 |
PS | P(0,0) and S(a,0) | PS= (a−0)2+(0−0)2 | PS=a |
PQ≅SRandQR≅PS
This theorem can also be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR.
PQ≅SRandQR≅PS