Circumradius of a `Delta ABC` is 2, O is the circumcentre, H is the orthocentre then- `1/64(AH^2+BC^2)(BH^2+AC^2)(CH^2+AB^2)` is equal to

In a Delta"le" ABC R=3 (circum radius) Let O be the circum centre and H be the ortho centre then (1)/(64)(AH^(2)+BC^(2))(BH^(2)+AC^(2))(CH^(2)+AB^(2))=

Circum radius of a DeltaABC is 3 units, let O be the circum and H be the orthocentre then the value of (1)/(64)(AH^(2)+BC^(2))(BH^(2)+AC^(2))(CH^(2)+AB^(2)) equals :

Circum radius of a DeltaABC is 3 units, let O be the circum and H be the orthocentre then the value of (1)/(64)(AH^(2)+BC^(2))(BH^(2)+AC^(2))(CH^(2)+AB^(2)) equals :

In Delta ABC is H is the orthocentre of Delta ABC and x= AH, y=BH, z=CH, then x+y+z=

In a Delta ABC the equation of the side BC is 2x-y =3 and its circumcentre and orthocentre are (2,4) and (1,2) respectively . Circumradius of Delta ABC is

In a Delta ABC the equation of the side BC is 2x-y =3 and its circumcentre and orthocentre are (2,4) and (1,2) respectively . Circumradius of Delta ABC is

In a Delta ABC the equation of the side BC is 2x-y =3 and its circumcentre and orthocentre are (2,4) and (1,2) respetively . The distance of orthocentre from vertex A is