MHB Show Diagonals of Kite are Perpendicular

[brinlin]

 

[HOI]

"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..

 

[Opalg]

View attachment 11290

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 $(\v c - \v a).\v d = \v 0$.

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.