Question

ABCD is rectangle. Diagonals intersect each other at point P. M is
is any point in the interior of rectangle then prove that : - MA²+MB²+ MC² + MD² = AC² + 4MP²



​

Answer 1

Answer:

Given the parallelogram is a rectangle.

In ∆OBC

angle BOC=30°(vertically opposite)

OB=OC(diagonals are equal and bisect each other)

So, using isoceles triangle property we have

angle 4OBC= angle OCB

Let it be x

so, by angle sum property

30°+x+x=180°

=>2x=180°−30=150°

x=150/2

x=75°

Now we know that each angle of a rectangle is 90°.

So we have angle OCD+OCB=90°

and angle OCB=75°

so, OCD+75=90°

OCD=90−75

OCD=15°

So we have angle OCD=15°