We want to prove one of the theorems already mentioned in the book using a coordinate proof. In a coordinate proof we place geometric figures in a coordinate plane. Then, we use variables to represent the coordinates of the figure.
Choosing a Theorem
We will choose a theorem that uses the lengths of a figure. A good example of this type of theorem is the Base Angles Theorem.
On a coordinate plane, let's consider a triangle with two congruent sides. For simplicity, we will plot vertex C on the y-axis. Therefore, its coordinates will be (0,c). Also, we will plot vertex B on the x-axis. This means that its coordinates will be (b,0). For congruency reasons, the coordinates of A will be (- b,0). Note that b and c are positive numbers!
We will now denote the origin (0,0) as H and consider the triangles △ AHC and △ BHC.
Let's now find the lengths of AH and BH. To do so, we will use the Distance Formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
Segment
Points
Substitute
Simplify
AH
A( - b, 0) and H( 0, 0)
AH = sqrt(( 0-( - b))^2+( 0- 0)^2)
AH=b
BH
B( b, 0) and H( 0, 0)
BH = sqrt(( 0- b)^2+( 0- 0)^2)
BH=b
We found that the distance from both A(- b,0) and B(b,0) to H(0,0) is equal to b. By the Transitive Property of Equality, this means that these distances are equal.
AH= b BH= b ⇒ AH=BH
By the definition of congruent segments, we know that AH≅HB. Additionally, note that by the Reflexive Property of Congruence HC is congruent to itself. Let's show this information in our diagram.
We found that △ AHC and △ BHC have three pairs of congruent sides.
AC≅ BC
AH≅ HB
HC≅ HC
By the Side-Side-Side Congruence Theorem, we can conclude that △ AHC ≅ △ BHC. Since corresponding parts of congruent figures are congruent, we know that ∠ HAC ≅ ∠ HBC. This proves the theorem.
