Answer:
Theorem: If a quadrilateral is a kite then it's diagonals are perpendicular.
Proof:- In kite WXYZ, let O(a,b) be the intersection point of the diagonals .
Also the distance formula is given by ,
Distance from \((x_1,y_2)\ and\ (x_2,y_2)=\sqrt{(x_2-x_1)^2-(y_2-y_1)^2}\)
In ΔWOX,
OX=a [since the coordinate of x changes but not y, thus the distance from O to X=a]
WO= 3b [since the coordinate of y changes but not x, thus the distance from O to W = 4b-b=3b]
WX=\(\sqrt{(2a-a)^2+(4b-b)}=\sqrt{a^2+9b^2}\)
We can see that
\(WX^2=a^2+9b^2=OX^2+WO^2\\\Rightarrow\ WX^2=OX^2+WO^2\)
\(\\\Rightarrow\ \angle{WOX}=90^{\circ}\) [By the converse of Pythagoras theorem]
⇒ The diagonals are perpendicular.
Hence proved.
