In Figure, `P` is a point in the interior of a parallelogram `A B C D` . Show that `a r( A P B)+a r( P C D)=1/2a r(^(gm)A B C D)` `a R(A P D)+a r( P B C)=a r( A P B)+a r( P C D)`

Explanation

If a triangle and parallelogram are on the same base and between the same parallel lines, then the area of the triangle will be half of the area of the parallelogram.
i) Let us draw a line segment `EF`, passing through the point P and parallel to line segment `AB` in parallelogram `ABCD`
.`AB || EF` (By construction) .....(1)

We know that `ABCD` is a parallelogram.
∴ `AD || BC` (Opposite sides of a parallelogram are parallel)
` AE || BF` ..... (2)
From Equations (1) and (2), we obtain
`AB || EF and AE || BF`
Therefore, quadrilateral `ABFE` is a parallelogram.
Similarly, it can be deduced that quadrilateral `EFCD` is a parallelogram.It can be observed that `triangleAPB` and parallelogram `ABFE` is lying on the same base `AB` and between the same set of parallel lines `AB` and `EF`
Area`(triangleAPB)` = 1/2 Area `(ABFE)` .....(3)
Similarly, for `trianglePCD` and parallelogram `EFCD`, Area `(trianglePCD)` = 1/2 Area `(EFCD)` .....(4)

Adding Equations (3) and (4), we obtain Area `(triangleAPB)` + Area `(trianglePCD)` = 1/2 [Area `(ABFE)` + Area `(EFCD)`]
Area `(triangleAPB)` + Area `(trianglePCD)` = 1/2 Area `(ABCD)` .....(5)

ii) Let us draw a line segment `MN`, passing through point P and parallel to line segment `AD`.
In parallelogram `ABCD`, `MN || AD` (By construction) .....(6)

We know that `ABCD` is a parallelogram.
`AB || DC` (Opposite sides of a parallelogram are parallel)
AM || DN.....(7)

From Equations (6) and (7), we obtain`MN || AD` and `AM || DN`
Therefore, quadrilateral `AMND` is a parallelogram.
It can be observed that `triangleAPD` and parallelogram `AMND` is lying on the same base `AD` and between the same parallel lines `AD` and `MN`.
Area `(triangleAPD)` = 1/2
Area (AMND) .....(8)
Similarly, for `trianglePCB` and parallelogram`MNCB`
Area `(trianglePCB)` = 1/2 Area `(MNCB)` .....(9)
Adding Equations (8) and (9), we obtain,
Area `(triangleAPD)` + `Area (trianglePCB)` = 1/2 [Area `(AMND)` + Area `(MNCB)`]
Area `(triangleAPD)` + Area `(trianglePCB)` = `1/2 `Area `(ABCD)` .....(10)

On comparing equations (5) and (10), we obtain
Area `(triangleAPD)` + Area `(trianglePBC)` = Area `(triangleAPB)` + Area`(trianglePCD)`