Question

In given fig. OB is perpendicular bisector of the line segment DE . FA is perpendicular to OB and FE intersects OB at point C. prove that 1/OA +1/OB = 2/OC






D F B O C A E

Answer 1

we have to use similar triangles! 

we can claim TRIAngle CAF ~ TRIangle CBE, 
and make this proportion: 

FA/OA = BE/BC AND also another similar triangles: TRI OAF ~ TRI OBD, which gives us: 
FA/OA = BD/OB 

but we know BC = OB - OC and AC = OB - OA - BC = OB - OA - OB + OC = OC - OA 

subbing back in, we get: 
OA/OB = (OC - OA)/(OB - OC) 
OA(OB - OC) = OB(OC - OA) 
OAOB - OAOC = OAOC - OAOB 

2OAOB = OAOC + OBOC 

now if we divide through by OAOBOC, we get what you're looking for: 

2/OC = 1/OB = 1/OA 
hope it helps you

Answer 2

Answer:

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