point inside a rectangle ABCD join to the vertices prove that the sum of areas of a pair of opposite triangle so formed is equal to the sum of areas of other pair of triangles
Answers
Answer:
So we have a rectangle ABCD
Point O is inside the rectangle where all the Vertices meet
In Triangles ACD and BCD
Angles 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 BCD
ACD - OCD = BCD - OCD (By Subtracting OCD from both the triangles)
We get,
AOC = BOD (proved)....(i)
In Triangles BAC and ACD
AB = 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)
=2AOC
From (iii)
AOC = AOB2AOC = 2AOB
RHS
AOB + AOB (SInce OCD =AOB)=2AOB
So we have a rectangle ABCD
Point O is inside the rectangle where all the Vertices meet
In Triangles ACD and BCD
Angles 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 BCD
ACD - OCD = BCD - OCD (By Subtracting OCD from both the triangles)
We get,
AOC = BOD (proved)....(i)
In Triangles BAC and ACD
AB = 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)
=2AOC
From (iii)
AOC = AOB2AOC = 2AOB
RHS
AOB + AOB (SInce OCD =AOB)=2AOB