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ICSE Class 9 Answered

ABCD is a parallelogram. A straight line is drawn Parallel to Diagonal BD,cutting BC and CD in P and Q respectively. Prove that triangle ABP and Triangle ADQ are equal in area
Asked by 742504142796845 | 22 Dec, 2018, 10:40: AM
answered-by-expert Expert Answer

Figure shows the parallelogram ABCD, line PQ dran parallel to BD , ΔABP and ΔADQ as given in the question.
let us draw line QR || BC and line SP || AB.
 
in ΔCDB, QP is parallel to DB;  hence begin mathsize 12px style fraction numerator C Q over denominator C D end fraction space equals space fraction numerator C P over denominator C B end fraction end style ..................(1)
Above eqn.(1) can be written as,
 
begin mathsize 12px style 1 minus fraction numerator C Q over denominator C D end fraction space equals space 1 space minus space fraction numerator C P over denominator C B end fraction space space space space space o r space space space space fraction numerator C D minus C Q over denominator C D end fraction space equals space fraction numerator C B minus C P over denominator C B end fraction space space h e n c e space fraction numerator D Q over denominator C D end fraction equals fraction numerator B P over denominator B C end fraction space equals space k end style
 
Let H and h are respective perpendicular height of parallelogram ARQD and ABPS
 
Area of paralleleogram ARQD = AR×H = AR×RQ×sinθ = DQ×BC×sinθ = k×CD×BC×sinθ .................(1)

Area of paralleleogram ABPS = AB×h = AB×BP×sinθ = AB×BP×sinθ = k×BC×AB×sinθ .................(2)

From (1) and (2), since AB = CD,   Area of paralleleogram ARQD = Area of paralleleogram ABPS  ....................(3)
 
Eqn.(3) is written as,   2×area of ΔADQ = 2×area of ΔABP 
 
hence , area of ΔADQ =  area of ΔABP
Answered by Thiyagarajan K | 27 Dec, 2018, 11:36: AM

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