Exercise 42 Page 510

We are asked to prove that the diagonals of a rectangle are congruent. We are asked to write a coordinate proof, so let's place the rectangle in a coordinate plane so that one vertex is at the origin and two sides are on the coordinate axes.

A rectangle is a parallelogram, so the opposite sides are parallel. Let's express the coordinates of C in terms of the coordinates of the other vertices.

• Point C is on a vertical line passing through D, so its x-coordinate is d.
• Point C is on a horizontal line passing through B, so its y-coordinate is b.

  To show that the diagonals are congruent, let's compare their lengths.

  
  We can use the Distance Formula to express these lengths using the coordinates of the endpoints. The distance between points(x_1,y_1)and(x_2,y_2)is sqrt((x_2-x_1)^2+(y_2-y_1)^2). Let's write and simplify an expression for the length of AC first.
  AC=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

  SubstitutePoints

  Substitute ( 0,0) & ( d,b)

  AC=sqrt(( d- 0)^2+( b- 0)^2)

  SubTerms

  Subtract terms

  AC=sqrt(d^2+b^2)
  Next, let's write and simplify an expression for the length of BD.
  BD=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

  SubstitutePoints

  Substitute ( 0,b) & ( d,0)

  BD=sqrt(( d- 0)^2+( 0- b)^2)

  SubTerms

  Subtract terms

  BD=sqrt(d^2+(- b)^2)

  NegBaseToPosBase

  (- a)^2=a^2

  BD=sqrt(d^2+b^2)
  We can see that AC=BD, so the diagonals are indeed congruent.