A point `O` in the interior of a rectangle `A B C D` is joined with each of the vertices `A ,\ B ,\ C` and `D` . Prove that `O B^2+O D^2=O C^2+O A^2`

A point O in the interior of a rectangle ABCD is joined with each of the vertices A,B,B,C and D. Prove that OB^(2)+OD^(2)=OC^(2)+OA^(2)

O is any point inside a rectangle ABCD. Prove that O B^2+O D^2=O A^2+O C^2 .

O is any point inside a rectangle ABCD. Prove that O B^2+O D^2=O A^2+O C^2 .

From a point O in the interior of a A B C , perpendiculars O D ,\ O E and O F are drawn to the sides B C ,\ C A and A B respectively. Prove that: (i) A F^2+B D^2+C E^2=O A^2+O B^2+O C^2-O D^2-O E^2-O F^2 (ii) A F^2+B D^2+C E^2=A E^2+C D^2+B F^2

In fig., O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:- O A^ 2 + O B^ 2 + O C^ 2 − O D^ 2 − O E^ 2 − O F^ 2 = A F^ 2 + B D^ 2 + C E^ 2

In figure, O is a point in the interior of a triangle ABC, O D_|_B C ,O E_|_A C and O F_|_A B . Show that (i) O A^2+O B^2+O C^2-O D^2-O E^2-O F^2=A F^2+B D^2+C E^2 (ii) A F^2+B D^2+C E^2=A E^2+C D^2+B F^2

In figure, O is a point in the interior of a triangle ABC, O D_|_B C ,O E_|_A C and O F_|_A B . Show that (i) O A^2+O B^2+O C^2-O D^2-O E^2-O F^2=A F^2+B D^2+C E^2 (ii) A F^2+B D^2+C E^2=A W^2+C D^2+B F^2

If O is the circumcentre of /_\ A B C and O D_|_B C , prove that /_B O D=/_A

A point O inside a rectangle A B C D is joined to the vertices. Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the sum of the other pair of triangles. Given: A rectangle A B C D\ a n d\ O is a point inside it. O A ,\ O B ,\ O C\ a n d\ O D have been joined. To Prove: a r\ (A O D)+\ a r\ ( B O C)=\ a r\ ( A O B)+\ a r( C O D)

A point O inside a rectangle A B C D is joined to the vertices. Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the sum of the other pair of triangles. Given: A rectangle A B C D\ a n d\ O is a point inside it. O A ,\ O B ,\ O C\ a n d\ O D have been joined. To Prove: a r\ (A O D)+\ a r\ ( B O C)=\ a r\ ( A O B)+\ a r( C O D)