Mathematics
In the adjoining figure, ABCD is a parallelogram. AB is produced to a point P and DP intersects BC at Q. Prove that : ar (ΔAPD) = ar (quad. BPCD).
Theorems on Area
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Answer
From figure,
AB // DC thus, AP // DC since AP is a straight line.
In parallelogram ABCD, the diagonal BD divides it into two triangles of equal area.
ar (ΔABD) = ar (ΔDBC) …..(1)
△ BPD and △ BPC lie on the same base BP and along the same parallel lines AP and DC.
ar (ΔBPD) = ar (ΔBPC) ….(2)
From figure,
ar (ΔAPD) = ar (ΔABD) + ar (ΔBPD)
Substituting values from equations (1) and (2) in above equation, we get :
ar (ΔAPD) = ar (ΔDBC) + ar (ΔBPC) ….(3)
From figure,
ar(quad. BPCD) = ar(△DBC) + ar(△BPC) ….(4)
From equations (3) and (4), we get :
∴ ar (ΔAPD) = ar(quad. BPCD)
Hence, proved that ar (ΔAPD) = ar(quad. BPCD).
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