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

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

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

In  Δ AOC and Δ COP

∠ AOC = ∠ COP    (common angle)

OA = OP      ( ∵ BP ║ AC  and O is intersection point of these lines )

OB = OC      ( ∵ BP ║ AC and O is intersection point of these lines )

∴ ar (Δ AOC) = ar (Δ COP)  (using SAS theorem)

Hence it is proved that, area of triangle AOB=area of triangle COP.

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

Given:

ABCD is a quadrilateral

BP ║ AC

Proof:

In Δ ABC and Δ PAC

CP = CP  (common base)

AC = AC (common side)

∴ ar (Δ ABC) = ar (Δ PAC)       (using SS criteria)

adding ar (Δ ADC) on both sides, we get,

ar (Δ ADC) + ar (Δ ABC) = ar (Δ ADC) + ar (Δ PAC)

as ADC + ABC = ABCD

and ADC + PAC = APD

we have,

∴ ar (ABCD) = ar (Δ APD)

Hence it is proved that, area of quadrilateral ABCD = to area of triangle APD​