Aditya Gupta
Last Activity: 5 Years ago
let coordinates of A be (a, 0) and B be (0, b). drop a perpendicular from D to the x axis and let it meet at P. let O be the origin.
then, angle AOB= angle DPA= 90 deg,
AB= DA (ABCD is square)
angle OBA= angle PAD ( because OAB+OBA= 90 and OAB+90+PAD= 180).
by AAS, triangles AOB and DPA are congruent. so, PD= OA= a
and PA= OB= b
so coordinates of D are (OA+AP, PD)= (a+b, a)
now, mid point of B and D is the centre of square.
by mid point theorem,
M= [(0+a+b)/2, (b+a)/2]
M= [(b+a)/2, (b+a)/2]= (h, k) say
so, h= (b+a)/2 and k= (b+a)/2
or h=k
replace h and k by x and y
x= y
x – y= 0
kindly approve :=)