Given points are A(-5,4), B(-1,-2) and C(5,2) are given.

Since these are vertices of an isosceles triangle ABC then AB = BC.

By distance formula, d(AB) = √[(x₂-x₁)²+(y₂-y₁)²]

Here x₁ = -5, y₁ = 4

x₂ = -1, y₂ = -2

d(AB) = √[(-1-(-5))²+(-2-4)²]

d(AB) = √[(4)²+(6)²]

d(AB) = √(16+36)

d(AB) = √52 …(i)

By distance formula, d(BC) = √[(x₂-x₁)²+(y₂-y₁)²]

Here x₁ = -1, y₁ = -2

x₂ = 5, y₂ = 2

d(BC) = √[(5-(-1))²+(2-(-2))²]

d(BC) = √[(6)²+(4)²]

d(BC) = √(36+16)

d(BC) = √52 …(ii)

From (i) and (ii) AB = BC

So given points are the vertices of isosceles triangle.

By distance formula, d(AC) = √[(x₂-x₁)²+(y₂-y₁)²]

Here x₁ = -5, y₁ = 4

x₂ = 5, y₂ = 2

d(AC) = √[(5-(-5))²+(2-4)²]

d(AC) = √[(10)²+(-2)²]

d(AC) = √(100+4)

d(AC) = √104 ….(iii)

Apply Pythagoras theorem to triangle ABC

AB²+BC² = (√52)²+(√52)²

= 52+52

= 104 ….(iv)

AC² = (√104)² = 104…(v)

From (iv) and (v) we got

AB²+BC² = AC²

So Pythagoras theorem is satisfied.

So the triangle is an isosceles right angled triangle.

Hence proved.

If ABCD is a square, let the diagonals meet at O.

Diagonals of a square bisect each other. So, O is the midpoint of AC and BD.

Consider A-O-C

x₁ = -5, y₁ = 4

x₂ = 5, y₂ = 2

By midpoint formula, x = (x₁+x₂)/2

x = (-5+5)/2 = 0/2 = 0

By midpoint formula, y = (y₁+y₂)/2

y = (4+2)/2 = 6/2 = 3

So co-ordinate of O is (0,3).

Consider B-O-D

Let co-ordinate of D be (x₂,y₂)

x₁ = -1, y₁ = -2

x = 0, y = 3

By midpoint formula, x = (x₁+x₂)/2

0 = (-1+x₂)/2

-1+x₂ = 0

x₂ = 1

By midpoint formula, y = (x₁+x₂)/2

3 = (-2+y₂)/2

6 = -2+y₂

y₂ = 6+2 = 8

Hence the co-ordinates of the point D are (1,8).