We are asked to prove that the diagonals of a rhombus are perpendicular. We will write a coordinate proof.

Since a rhombus is a parallelogram, we can use the following coordinates for the vertices.

We start with $A (0, 0),$ $B (a, b),$ and $D (c, 0) .$
The $y -$ coordinate of $C$ is set as the $y -$ coordinate of $B$ so that $\overline{B C}$ is parallel to the $x -$ axis.
The $x -$ coordinate of $C$ is set so that $\overline{B C}$ and $\overline{A D}$ have the same (horizontal) length.

Let's use the Distance Formula to find the length of

\overline{A B}

and

\overline{A D} .

A B A D = (a - 0)^{2} + (b - 0)^{2} = a^{2} + b^{2} = (c - 0)^{2} + (0 - 0)^{2} = c^{2} = c

Since

A B C D

is a rhombus, we know that

A B = A D .

This gives us a relationship between the variables

a,

b,

and

c .

a^{2} + b^{2} = c ⟹ a^{2} + b^{2} = c^{2}

We are asked to investigate the direction of the diagonals, so let's use the Slope Formula to find the slopes.

We are asked to show that the diagonals are perpendicular. This is true if the slopes multiply to

- 1,

so let's multiply the expressions above and simplify the product.

m_{A C} m_{D B} = \frac{b}{a + c} \cdot \frac{b}{a - c}

Multiply fractions

m_{A C} m_{D B} = \frac{b ^{2}}{( a + c ) ( a - c )}

$(a + b) (a - b) = a^{2} - b^{2}$

m_{A C} m_{D B} = \frac{b ^{2}}{a ^{2} - c ^{2}}

We know from above that

a^{2} + b^{2} = c^{2} .

This can be rearranged to

a^{2} - c^{2} = - b^{2},

so let's substitute this in the denominator.

m_{A C} m_{D B} = \frac{b ^{2}}{a ^{2} - c ^{2}}

$a^{2} - c^{2} = - b^{2}$

m_{A C} m_{D B} = \frac{b ^{2}}{- b ^{2}}

Put minus sign in front of fraction

m_{A C} m_{D B} = - \frac{b ^{2}}{b ^{2}}

$\frac{a}{a} = 1$

m_{A C} m_{D B} = - 1

Since the product of the slopes is

- 1,

diagonals

\overline{A C}

and

\overline{D B}

of rhombus

A B C D

are indeed perpendicular. This finishes the coordinate proof of Theorem

6 - 13

.

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