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)
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 $\displaystyle (\overline{PA})^2 + (\overline{PB})^2 = (\overline{PC})^2 + (\overline{PD})^2$