Exercise 28 Page 501

We are asked to show the following statement. Ifboth pairs of opposite sides of a quadrilateral are congruent, thenthe quadrilateral is a parallelogram. We are asked to write a coordinate proof, so place the quadrilateral in the coordinate plane. It is helpful if you choose a position with one vertex at the origin and one side on the positive x-axis. Let's label the coordinates of the vertices of the quadrilateral.

Segments are congruent if they have the same length, so let's recall the Distance Formula. The distance between pointsP(x_1,y_1)andQ(x_2,y_2)is PQ=sqrt((x_2-x_1)^2+(y_2-y_1)^2). Let's use this formula to set up equations that express that the opposite sides of the quadrilateral are congruent. Instead of the length of the segments, we can set up the equations for the squares of the segments.

• Segments AD and BC are congruent, so AD^2=BC^2.

(b-0)^2+(c-0)^2=(x-a)^2+(y-0)^2

• Segments AB and DC are congruent, so AB^2=DC^2.

(a-0)^2+(0-0)^2=(x-b)^2+(y-c)^2 After simplification, this gives us the following system of equations. b^2+c^2=(x-a)^2+y^2 a^2=(x-b)^2+(y-c)^2 A solution of this system is (x,y)=(a+b,c). We can check this by substitution.

This substitution confirms the fourth vertex of the quadrilateral in terms of the coordinates of the other three vertices.

To show that this quadrilateral is a parallelogram, we need to prove that opposite sides are parallel.

• Points D and C have the same second coordinates, so DC is parallel to the x-axis. This means that DC and AB are parallel.
• To get from A(0,0) to D(b,c), we need to move b units to the right and c units up. This is the same movement as the movement from B(a,0) to C(a+b,c). This means that AD and BC are parallel.

Since opposite sides of ABCD are parallel, by definition, this is a parallelogram.

Extra

Solving the equation system

In the solution we set up the following system of equations. b^2+c^2=(x-a)^2+y^2 a^2=(x-b)^2+(y-c)^2 Without solving this system, we stated the solution (x,y)=(a+b,c) and checked it. If you solve this system, you find that there is another solution. x&=(a-b)(a^2-b^2-c^2)/(a-b)^2+c^2 y&=-c(a^2-b^2-c^2)/(a-b)^2+c^2 This second solution also gives a point at the correct distance from D and B, but in this case ABCD is not a quadrilateral.