Question

In the given figure OB is perpendicular birector of the line segment DE, FAOB and
FE intersects Os at the point C. Prove that​

Answer 1

Step-by-step explanation:

OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure .

now, ∆OAF and ∆ODB

∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF }

∠FOA = ∠DOB { common angle }

from A - A similarity , ∆OAF ~ ∆ODB

so, OA/OB = AF/DB = OF/OD -------(1)

similarly, ∆AFC and ∆BEC

∠FCA = ∠BCE

∠FAC = ∠CBE = 90°

from A - A similarity , ∆AFC ~ ∆BEC

so, AF/BE = AC/CB = FC/CE -------(2)

we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2)

AF/DB = AC/CB =FC/CE -------(3)

now, from equations (1) and (3),

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

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

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

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

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

dividing by OA.OB.OC both sides,

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

2/OC = 1/OA + 1/OB

hence, \bold{\frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}}hence,

OA

1

+

OB

1

=

OC

2