(b) ABCD is a parallelogram. BP and DQ are the perpendiculars from B and D on the diagonal AC.
(i) Show that BP = DQ.
(ii) Show that ar ( ADX) = ar ( ABX), where X is any point in AC. 

Dear Student,

Consider the triangles , BCA and DACAB=CD              opposite sides of parallelogramAD=BC              opposite sides of parallelogramAC=AC              CommonHence BCA DAC   By SSSSo,i The area of the corresponding triangles are also equal arBCA =arDAC 12×DQ×AC=12×BP×ACHence BP=DQ   provediiNow consider the triangle ADX andABXAX is the base common to both the triangles  ADX andABXAs, Area of the triangle ADX=12×AX×DQ      DQ is the altitudeSimilarly , Area of the triangle ABX=12×AX×BP      BP is the altitudeSince BP=DQ from part iwe have, 12×AX×DQ=12×AX×BP So, arADX=arABX  Proved
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