# Solve this: ​Q24. In the given figure, ABCD is a parallelogram. AB is produced to a point P and DP intersects BC at Q. Prove that area ($∆$APD) = area (quad. BPCD)    [Hint : Diagonal divides a || gm into two triangles of equal area $\therefore$ area ​​$∆$ABD = area ​$∆$DBC     Again, area ​$∆$BPD = area ​$∆$BPC ($\because$ Triangles on the same base and between the same parallel are equal in area )

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