# Prove the diagonals of a kite are perpendicular.

1. Oct 10, 2011

### MklStar

1. The problem statement, all variables and given/known data
As the title says, the question asks of us to: Prove that the diagonals of a non square kite are perpendicular using vectors.

2. Relevant equations
if a is perpendicular to b then a . b =0

3. The attempt at a solution
a kite made from points A B C D
..............._a_A
.............b| /\
.......D.....|/....\......B
...........c|.\..../
.............|...\/
C
DC = a + c
b≠c
i know we have to use dot products but im not sure how to implicate that into this...

2. Oct 10, 2011

### daveb

Try drawing the sides as vectors and the diagonals as the vector sum of two sides.

3. Oct 10, 2011

### HallsofIvy

Staff Emeritus
A "kite" is a four sided figure having two pairs of congruent sides in which the congruent sides are adjacent. One "diagonal" separates the figure so that two congruent sides are one side of the diagonal, the other two congruent sides on the other. That is, you have two isosceles triangles. An altitude of an isosceles triangle, to the non-congurent side, is the perpendicular bisector of that side.

4. Oct 11, 2011

### MklStar

I get what you mean but how would I explain that with vectors?
I dont think I can just split the kite into 2 isosceles triangles and just say that from the middle of the base to the top, the angle is perpendicular.

5. Oct 11, 2011

### daveb

As I said, draw the kite with the sides as vectors. Each diagonal is the vector sum of two sides. Take the dot product of those two diagonals (written as sums) and you should see how they work out to equal zero.

6. Oct 11, 2011

### HallsofIvy

Staff Emeritus
You can always assume a coordinate system so that one vertex, where two congruent sides join, is the origin and the diagonal at that point is the y-axis. Then the other vertex on that diagonal is (d, 0) for some number d and the other two points are (a, 0) and (-a, 0) for some number a. Construct the vectors along the diagonals and take the dot product.