# In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.If BO = OD, prove thatAr(∆ABC) = ar(∆ADC). Given : A quadrilateral ABCD with diagonals AC and BD and BO = OD

To prove: Area of (∆ABC) = area of (∆ADC)

Proof : BO = OD [given]

Here AO is the median of ∆ABD Area of (∆AOD) = Area of (∆AOB) ---------------- 1

And OC is the median of ∆BCD Area of (∆COD) = Area of (∆BOC) ---------------- 2

Now by adding –1 and –2 we get

Area of (∆AOD) + Area of (∆COD) = Area of (∆AOB) + Area of (∆BOC) Area of (∆ABC) = Area of (∆ADC)

Hence proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Quiz | Area and Parallelogram46 mins  Champ Quiz | Area of Triangle53 mins  Quiz | Surface Area & Volumes49 mins  Surface Area of Right Circular Cylinder52 mins  Quiz | Mensuration42 mins  Champ Quiz | Area of Parallelogram55 mins  Surface Area and Volume of Cone24 mins  Quiz | Surface Area and Volumes43 mins  Surface Area of Right Circular Cylinder49 mins  Surface Area of Cube and Cuboid49 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 