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`

In A B C ,\ \ /_A=60o . Prove that B C^2=A B^2+A C^2-A B.A C .

The diagonals of a parallelogram A B C D intersect at OdotA line through O intersects A B\ at X\ a n d\ D C at Ydot Prove that O X=O Y

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 W^2+C D^2+B F^2

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 and O is a point inside it. O A ,O B ,O C and 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) CONSTRUCTION : Draw E O F A B and L O M A Ddot

O is any point on the diagonal B D of the parallelogram A B C Ddot Prove that a r( O A B)=a r( O B C)

In figure, ABC is triangle in which /_A B C > 90o and A D_|_C B produced. Prove that A C^2=A B^2+B C^2+2B C.B D .

The diagonals of a rectangle A B C D\ meet at O . If /_B O C=44^0, find /_O A D

In a quadrilateral A B C D , given that /_A+/_D=90o . Prove that A C^2+B D^2=A D^2+B C^2 .