# Circle

• Nov 29th 2010, 05:14 PM
prasum
Circle
A circle is inscribed into a rhombus abcd with one angle 60 degree.the distance from the centre of the circle to the nearest vertex is equal to 1.if p is any point on the circle then square(pa)+square(pb)+square(pc)+square(pd)
• Nov 29th 2010, 05:58 PM
Educated
Quote:

Originally Posted by prasum
A circle is inscribed into a rhombus abcd with one angle 60 degree.the distance from the centre of the circle to the nearest vertex is equal to 1.if p is any point on the circle then square(pa)+square(pb)+square(pc)+square(pd)

Um... what is the question?

Is it supposed to be: Prove that $(\overline{PA})^2 + (\overline{PB})^2 = (\overline{PC})^2 + (\overline{PD})^2$
• Nov 30th 2010, 01:34 AM
prasum
we have to find the value