Question

answer the following.
In the given figure OB is perpendicular bisector of the line segment DE, FA perpendicular to OB and
FE intersects OB at the point C. Prove that
$\frac{1}{ob} + \frac{1}{ oc} = \frac{2}{oc}$
​

Answer 1

Answer:

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

Now △OAF and △ODB

∠OAF=∠OBD=90

o

(because OB ⊥er bisector of DE,So OB⊥AF and OB⊥DE)

∠FOA=∠DOB (common angle)

From A−A, △OAF≅△ODB

Similarly △AFC and △BEC

∠FCA=∠BCE,∠FAC=∠CBE=90

o

from A−A, △AFC≅△BEC

So AF/BE=AC/CB=FC/CE

We know taht DE=BE (⊥ er 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

2OAOB=OB.OC+OA.OC

Divide by OA.OB.OC

OAOBOC

2OAOB

=

OAOBOC

OB.OC

+

OAOBOC

OA.OC

OC

2

=

OA

1

+

OB

1

∴

OA

1

+

OB

1

=

OC

2

hope it helps you