EXPERTS answer the question attached ......... Share with your friends Share 0 Atul Shaw answered this Dear student, Firstly draw two circles with center O and O’ such that they intersect at A and B.Draw a line PQ parallel to OO’. In the circle with center O, we have: OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector. i.e. BM=MP....(1) In the circle with center O’, we have: O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector. i.e. BN=NQ....(2)BM=MP....(3) From (1) and (2), we have: BM+BN=MP+NQ ⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ) ⇒2(BM+BN)=(BM+BN)+(MP+NQ) ⇒2(OO’)=(BM+MP)+(BN+NQ) ⇒2(OO’)=BP+BQ ⇒2OO’=PQ Hence, proved. Regards 0 View Full Answer Bhavika ... answered this Don't dlt 0 Itz Vruta 😘🖤🖤🖤 answered this Okkkk 0 Itz Vruta 😘🖤🖤🖤 answered this Good morning 0 Bhavika ... answered this Good morning... 0 Itz Vruta 😘🖤🖤🖤 answered this Hru... 0 Bhavika ... answered this Good , u? 0 Itz Vruta 😘🖤🖤🖤 answered this Fantastic 0
