Theorem: Diagonals of Kite Intersect at Right Angles
The interesting properties of the kite is that its diagonal are always perpendicular to each other. This is proved below, we have a kite ABCD, whose diagonal intersect each other at point O.
In ∆ABD and ∆BCD
AB = BC (Property of Kite)
AD = CD (Property of Kite)
BD = BD (Common Side)
Thus, ∆ABD ≅ ∆BCD (SSS congruency)
Now, in ∆ABC and ∆ADC
AB = BC (Property of Kite)
Hence ∆ABC is an isosceles triangle.
AD = CD (Property of Kite)
Hence ∆ADC is an isosceles triangle.
∠BAO = ∠BCO
BO = BO (Common Side)
Thus, ∆ABO ≅ ∆BCO (SAS rule of congruency)
Now we know ∠AOB = ∠BOC
Also, ∠AOB + ∠BOC = 180° (Linear Pair)
Hence, ∠AOB = ∠BOC = 90°
Hence diagonals of kite intersect at right angles.
