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Example (14: O is any point inside arectangle ABCD (see Fig. 6.52). Prove that |
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Answer» Ans :- We draw PQ ║AB ║CD as shown in figure, ABCD is a rectangle , it means ABPQ and PQDC are also rectangle For , ABPQ , AP = BQ [opposite sides are equal ] For, PQDC PD = QC [ opposite sides are equal ] now, for ∆OPD,OD² = OP² + PD² ------(1) For, ∆OQB, OB² = OQ² + BQ² --------(2) add equations (1) and (2), OB² + OD² = (OP² + PD²)+ (OQ² + BQ²) = (OP² + CQ²) + (OQ² + AP²) as you can see figure, ∆OPA and ∆OQC are also right angled triangles For, ∆OCQ ⇒OQ² + CQ² = OC²For,∆OPA ⇒OP² + AP² = OA² , put it above Now , OB² + OD² = OC² + OA² thanks |
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