Question

A point O inside a rectangle ABCD is joined to vertices.Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the other pair of triangles

Answer 1

So we have a rectangle ABCDPoint O is inside the rectangle where all the Vertices meetIn Triangles ACD and BCDAngles ACD= BDC = 90° (Angle of between two sides of a rectangle is  90  degree)AC = BD (Opposite sides are equal)CD=CD (Common)Triangles ACD is congruent to BCD (By Side Angle Side)Hence ,Area of ACD = Area of BCDACD - OCD = BCD - OCD  (By Subtracting OCD from both the triangles)We get,AOC = BOD (proved)....(i)
In Triangles BAC and ACDAB = CD (opposite sides are equal in a rectangle)Angles BAC = DCA = 90°Thus, Triangles BAC is congruent to ACD (by Side Angle Side)Areas of BAC = ACD BAC - AOC = ACD - AOC (Subtracting AOC from both triangles)AOB = OCD (proved)...(ii)Since,Triangles BAC = ACD = BCD (Proved earlier)AOC = BOD = AOB = OCD....(iii)AOC + BOD = AOB + OCD
LHS  AOC + AOC (Since AOC = BOD) =2AOCFrom (iii)AOC = AOB2AOC = 2AOB 
RHSAOB + AOB (SInce OCD =AOB)=2AOB

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Answer 2

plz mark brainliest if helped