Question

How do you Prove that the midpoints of the sides of a kite form a rectangle?

Answer 1

Conditions

How do you Prove that the midpoints of the sides of a kite form a rectangle?

Solution

Consider a graph:

As we know, a kite has a form of two isosceles triangles with a common base.

As WX=XY, and A and B are midpoints, then AB is parallel to WY. XO is a height, so angle XTB is 90 degrees. The angle XOY is 90 degrees too, as the diagonals of rhombus are intersecting with the right angle.

Consider the triangle XOY:

$$\angle X O Y = 9 0 {}^ {\circ}; \angle O Y X = 9 0 {}^ {\circ} - \angle O X Y$$ $$\angle O Y X = \angle A B X, \text{because } A B \text{ and } W Y \text{ are parallel.}$$

As B and C are midpoints, then XZ is parallel to BC, then:

$$\angle O Y B = \angle O X Y$$

The angle

$$\begin{array}{l} \angle X B Y = 1 8 0 {}^ {\circ} = \angle X B A + \angle A B C + \angle O B Y = \angle O Y X + \angle A B C + \angle O X Y \\ = \angle A B C + 9 0 {}^ {\circ} - \angle O X Y + \angle O X Y = \angle A B C + 9 0 {}^ {\circ} \\ \end{array}$$ $$1 8 0 {}^ {\circ} = \angle A B C + 9 0 {}^ {\circ}$$ $$\angle A B C = 9 0 {}^ {\circ}$$

Analogically we can consider other 3 triangle and prove that each angle of ABCD is $90{}^{\circ}$. And this is a rectangle by the definition of rectangle.

Q.E.D.