Question

A point O inside a rectangle ABCD 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.

Draw Figure too! Thank you! ​

Answer 1

Given :

• A rectangle ABCD and O is a point inside it. OA,OB,OC and OD have been joined.

To Prove :

• ar(AOD) + ar(∆BOC) = ar(∆AOB) + ar(∆COD)

Construction :

• Draw EOF || AB

• LOM || AD

Proof :

We have,

ar(∆AOD) + ar(∆BOC) = ½(AD × BC) + ½(BC × OF)

→ ar(∆AOD) + ar(∆BOC) = ½(AD × OE) + ½(AD × OF)

$(\because\sf AD\: =\: BC)$

→ ar(∆AOD) + ar(∆BOC) = ½AD × (OE +OF)

→ ar(∆AOD) + ar(∆BOC) = ½(AD × EF)

→ ar(∆AOD) + ar(∆BOC) = ½(AD × AB)

$(\because\sf EF\: =\: AB)$

→ ar(∆AOD) + ar(∆BOC) = ½ ar(Rectangle ABCD)...i)

and,

→ ar(∆AOB) + ar(∆COD) = ½(AB×OL) + ½ (CD × OM)

→ ar(∆AOB) + ar(∆COD) = ½(AB×OL) + ½ (AB × OM)

$(\because\sf AB\: =\: CD)$

→ ar(∆AOB) + ar(∆COD) = ½(AB × (OL + OM)

→ ar(∆AOB) + ar(∆COD) = ½(AB × LM)

→ ar(∆AOB) + ar(∆COD) = ½(AB × AD)

$(\because\sf LM\: =\: AD)$

→ ar(∆AOB) + ar(∆COD) = ½ ar(Rectangle ABCD)....ii)

From i) and ii),we get :

ar(∆AOD) + ar(∆BOC) = ar(∆AOB) + ar(∆COD)

Hence,Proved!

Answer 2

$\huge\underline\red{Answer:-}$

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