Show Diagonals of Kite are Perpendicular

In summary: This approach is not correct. In summary, the conversation discusses the concept of diagonals in a kite and how to prove that they are perpendicular to each other. The approach of choosing points A, C, B, and D on the x and y axes is incorrect as it assumes the perpendicularity of the diagonals. Instead, the conversation suggests labeling the point of intersection of the diagonals as E and using vector representation to prove the perpendicularity of the diagonals.
  • #1
brinlin
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  • #2
"AB- BC"? Do mean "AB= BD"? Personally, I would NOT "let B be the origin". I would label the point where AC and BD intersect "E" and let that be the origin. Then A is (-a, 0) for some number, a, and C is (a, 0). B is (0, b) and D is (0,-d) (d is not necessarily equal to b). Then AB= <0, b>- <a, 0>= <a, b> and CB= <0, b>- <a, 0>= <-a, b> so that AB and CB have the same length, $\sqrt{a^2+ b^2}$. AD= <0, -d>- <-a, 0>= <a, -d> and CD= <0, -d>- <a, 0>= <a, -d> so that AD and CD also have the same length, $\sqrt{a^2+ d^2}$. Yes this is a "kite".

The diagonals are AC and BD. AC= <-a, 0>- <a, 0>= <-2a, 0>. BD= <0, b>- <0, d>= <0, b- d>. The dot product of AC and BD is <-2a, 0>.<0, b-d>= (-2a)(0)+ (0)(b-d)= 0. Therefore the diagonals are perpendicular..
 
  • #3
brinlin said:
Taking $B$ as the origin, let $A,C,D$ be represented by vectors $\def\v{\mathbf} \v a,\v c,\v d$. You are told that $AB = BC$, which says that $\v{a.a} = \v{c.c}$. Also, $CD = DA$, so that $(\v d - \v c)\v.(\v d - \v c) = (\v d - \v a)\v.(\v d - \v a)$. Using those equations, you want to show that $AC = BD$, or in other words $(\v c - \v a).\v d = \v 0$.
Country Boy said:
I would label the point where AC and BD intersect "E" and let that be the origin. Then A is (-a, 0) for some number, a, and C is (a, 0). B is (0, b) and D is (0,-d) ...
By choosing $A$ and $C$ to be on the $x$-axis, and $B$ and $D$ to be on the $y$-axis, you are assuming that $AC$ is perpendicular to $BD$, which is what you are supposed to be proving.
 

FAQ: Show Diagonals of Kite are Perpendicular

1. How do you prove that the diagonals of a kite are perpendicular?

The diagonals of a kite are perpendicular if they intersect at a right angle. This can be proven using the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. By applying this theorem to the two triangles formed by the diagonals of a kite, we can show that the diagonals are perpendicular.

2. What is a kite in geometry?

A kite is a quadrilateral with two pairs of equal adjacent sides. It is a type of quadrilateral known as a "special parallelogram" because it has two pairs of parallel sides, and its opposite angles are equal.

3. Why are the diagonals of a kite perpendicular?

The diagonals of a kite are perpendicular because they intersect at a right angle. This is a property of kites that can be proven using the Pythagorean Theorem.

4. Can you provide an example of a kite in real life?

Yes, a kite is a common shape found in everyday objects. For example, a diamond or rhombus-shaped sign is a real-life example of a kite. Another example is a kite used for flying, which has two pairs of equal adjacent sides and two perpendicular diagonals.

5. How is the property of perpendicular diagonals useful in geometry?

The property of perpendicular diagonals in a kite is useful in geometry because it can be used to solve problems involving angles and sides of a kite. It can also be applied to other shapes and figures, such as rectangles and squares, to determine if their diagonals are perpendicular.

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