Question

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☆Solve this :-》》
In the given figure, OB is perpendicular bisector of the segment DE, FA ..........
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Question no. 28 ☆☆☆☆☆
In *attachment*
Urgent please for board 10th

Answer 1

Wow! That's the exact question I was doing right now

Given that∠AEF=∠AFE and E is mid point of CA i.eCE=AE
Draw CG∥DF

Hence by Basic Proportionality theorem(BPT) we have BDCD=BFGF ;(i)In ∆AFE∠AEF=∠AFE⇒AF=AEAlso CE=AE⇒AF=AE=CE ;(ii)In ∆ACGE is the mid point of AC, and EF∥CG⇒FG=AF ;
(iii)By (i),(ii) and 
(iii), we haveBDCD=BFCE (Hence Proved)

Answer 2

Answer:

(1/OA) + (1/OB) = (2/OC)

Step-by-step explanation:

From Figure:

(i) In ΔAOF and ΔBOD.

⇒ ∠O = ∠O

⇒ ∠A = ∠B

⇒ ΔAOF ~ ΔBOD

⇒ OA/OB = FA/DB

(ii) In ΔFAC and ΔEBC,

⇒ ∠A = ∠B

⇒ ∠FCA = ∠ECB

⇒ ΔFCA ~ ΔEBC

⇒ FA/EB = AC/BC

But,

EB = DB

FA/DB = AC/BC

From (i) & (ii), we get

⇒ AC/BC = OA/OB

⇒ (OC - OA)/(OB - OC) = OA/OB

⇒ OB * OC - OA * OB = OA * OB - OA * OC

⇒ OB * OC + OA * OC = 2 OA * OB

⇒ OC(OB + OA) = 2 OA * OB

⇒ (OB + OA)/OA * OB = 2/OC

⇒ (1/OA) + (1/OB) = 2/OC.

Hope it helps!