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Abcd is a Parallelogram Whose Diagonals Intersect at O. If P is Any Point on Bo, Prove That: (I) Ar (δAdo) = Ar (δCdo) (Ii) Ar (δAbp) = Ar (δCbp) - Mathematics

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove
that:  (1) ar (ΔADO) = ar (ΔCDO)     (2) ar (ΔABP) = ar (ΔCBP)

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Solution

Given that ABCD is a parallelogram
To prove: (1) ar (ΔADO)  = ar (ΔCDO) 

  (2) ar ( ΔABP) = ar (ΔCBP)

Proof: We know that, diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
     
(1)    In ΔDAC , since DO is a median
Then area  (ΔADO) =  area (ΔCDO)
 
(2)   In ΔBAC ,  Since BO is a median
Then; area (ΔBAO)  area (ΔBCO)    ......(1)
In a ΔPAC, Since PO is a median
Then, area (ΔPAO) = area (ΔPCO)   ......(2)
Subtract equation (2) from equation (1)
 
⇒  area (ΔBAO) - ar (ΔPAO)  ar (ΔBCO) - area (ΔPCO)
⇒  Area (ΔABP) = Area of ΔCBP
 
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 12 | Page 46
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