Question

this question is very hard no one can do this. ​

Answer 1

OB is perpendicular bisecter of line segment DE,FA perpendicular to DB and FE intersect OB at the point C as shown in figure

Now ∆ OAF and ∆ ODB

Angel OAF =Angel ODB=90deegre

(because DB perpendicular bisector of DE , so OB perpendicular AF and OB perpendicular DE) angel FOA = angel DOB (common angel )

From A-A , ∆ OAF congruent ∆ ODB

Similarly ∆ AFC and ∆ BEC

angel FCA =angel, angel FAC = angel CBE = 90 degree

From A-A, ∆ AFC congruent ∆ BEC so

AF / BE= AC/CB = FC/CE

we know that DE= BE perpendicular bisector of DE is OB

AF / DB=AC/CB=FC/CE

as we have

OA/OB =AF/DB=OF/OD

OA/OB= AC/CB=(OC-OA) /(OB-OC )

OA/OB=(OC-OA) /(OB-OC)

OA(OB-OC)=(OC-OA)OB

OA.OB-OA.OC=OB.OC-OA.OB

2OA.OB=OB.OC+OA.OC

divide by OA.OB.OC

2OAOB/OA.OB.OC=OB.DC/OA.OB.OC +OA.OC/OA.OB.OC

2/OC=1/OA+1/OB

.: 1/OA+1/ OB=2/OC