Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB)×ar(CPD)=ar(APD)×ar(BPC)

Let the ∠APB be equal to α and ∠CPD be equal to β. So, we get ∠BPC is equal to 180-α and ∠APD is equal to 180-β.
We know that ar(ΔPQR) = (1/2)pq sin∠PRQ
A(ΔAPB) = (1/2)AP × BP × sinα
A(ΔBPC) = (1/2)CP × BP × sin(180-α) = (1/2)CP × BP × sin(α) [∵sin(180-α) = sinα]
A(ΔCPD) = (1/2)CP × DP × sin(β)
A(ΔAPD) = (1/2)AP × DP × sin(180-β) = (1/2)AP × DP × sin(β).
So,
A(ΔAPB) × A(ΔCPD) = A(ΔBPC) × A(ΔAPD)
⇒(1/2)AP × BP × sinα × (1/2)CP × DP × sinβ = (1/2)CP × BP × sinα × (1/2)AP × DP × sinβ
⇒AP × BP × CP × DP × sin²α = CP × DP × AP × DP × sin²β
This equation is not generally true. The correct relationship is derived from the areas of the triangles using the fact that triangles APB and CPD share the same angles, as do triangles APD and BPC. The areas are proportional to the product of the lengths of the sides forming the angle.
Let's reconsider using the area formula Area = (1/2)ab sin C
Area(APB) = (1/2) AP * BP * sin(∠APB)
Area(BPC) = (1/2) BP * CP * sin(∠BPC)
Area(CPD) = (1/2) CP * DP * sin(∠CPD)
Area(DPA) = (1/2) DP * AP * sin(∠DPA)
Since ∠APB = ∠CPD and ∠BPC = ∠DPA (vertically opposite angles), we have:
Area(APB) * Area(CPD) = (1/4) AP * BP * CP * DP * sin(∠APB) * sin(∠CPD) = (1/4) AP * BP * CP * DP * sin²(∠APB)
Area(BPC) * Area(DPA) = (1/4) BP * CP * DP * AP * sin(∠BPC) * sin(∠DPA) = (1/4) AP * BP * CP * DP * sin²(∠APB)
Therefore, Area(APB) * Area(CPD) = Area(BPC) * Area(DPA)