plz solve this problem plzzzzzzzzZ
Answers
Answer:
This can be proved by applying similarity of triangles .
Step-by-step explanation:
In ΔOAF and ΔOBD
∠O = ∠O (common)
∠OAE = ∠OBD (given 90° as perpendicular)
=> ΔOAF ~ ΔOBD [BY AA criteria]
=> OA/OB = FA/DB (sides are proportional of similar triangles) .........(1)
Also, in ΔFAC and ΔEBC
∠FAC = ∠EBC (given perpendicular)
∠FCA = ∠BCE (vertically opposite angles)
=> ΔFAC ~ ΔEBC [AA criteria]
Thus, FA/EB = AC/BC (sides are proportional of similar triangles)
But , EB = DB (given)
=> FA/DB = AC/BC ...................(2)
From equation(1) and (2)
OA/OB = AC/BC
=> OA/OB = (OC - OA)/(OB - OC)
=> OA * (OB - OC) = OB * (OC - OA)
=> OA * OB - OA * OC = OB * OC- OB * OA
=> OA * OB + OB * OA = OB * OC + OA * OC
=> 2(OA * OB) = (OA + OB) * OC
Now, Dividing both sides by OA * OB * OC
=> 2(OA * OB)/[OA * OB * OC] = [(OA + OB) * OC]/[OA * OB * OC]
=> 2/OC = [(OA + OB)]/[OA * OB]
=> 2/OC = OA/[OA * OB] + OB/[OA * OB]
=> 2/OC = 1/OB + 1/OA
=> 1/OB + 1/OA = 2/OC
Hence proved .