Question

In given figure, OB is the perpendicular bisector of the line segment $DE,FA\perp OB$ and $\stackrel{-}{F}\stackrel{-}{E}$ intersects $\stackrel{-}{O}\stackrel{-}{B}$ at point C. Prove that :
$\frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}$

Answer 1

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