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Question -

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



Answer -

Given:
(1) ABCD is a quadrilateral,
(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.
To prove: Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC
Construction: Draw AL perpendicular to BD and CM perpendicular to BD
Proof:
We know that
Area of triangle =  × base× height
Area of ÄAPD =  . DP . AL …… (1)
Area of ÄBPC =   . CM . BP …… (2)
Area of ÄAPB =   . BP . AL …… (3)
Area of ÄCPD =  . CM . DP …… (4)
Therefore
 
Hence it is proved that  

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