Question

in the given figure ABCD is a quadrilateral.BP is drawn parallel to AC and BP meets DC produced to P .prove that

1) area of triangle AOB=area of triangle COP.


2) area of quadrilateral ABCD = to area of triangle APD​

Answer 1

Given: A quadrilateral ABCD, BE is parallel to AC.

To find: Area of triangle AOB=area of triangle COP

             Prove that area (ADP) = area (ABCD)

             

Solution:

• To prove that  area (ADP) = area (ABCD), we first need to prove that area (ABC) = area (ACP),

• So, as we can see that, triangle BAC and triangle EAC lies on the same base AC, and lies between the same parallels, that is AC and BE.

• Therefore,

             area (BAC) = area (PAC)

• Now, subtract area (AOC) from both sides, we get:

              area (BAC) - area (AOC) = area (PAC)  - area (AOC)

              area (AOB) = area (COP)

• Now, we need to add area (ADC) on both sides, we get:

              area (BAC) + area (ADC) = area (EAC) + area (ADC)

• Now, we know that area (BAC) + area (ADC) = area (ABCD), so:

Answer:

      area (ABCD) = area (ADE) ....................hence proved.