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2015-04-01T12:13:04+05:30
given that , 
   ∠1 = ∠2
   OA = OD 
∠CAO = 180 - ∠1 ( linear pair ) 
 ∠BDO = 180 - ∠2 (  "      "  )
if , ∠1 = ∠2 
   ⇒ 
∠CAO = ∠BDO ------ ( i ) 

 now in , ΔCAO & Δ BDO 
               
 ∠CAO = ∠BDO  ( From eqn. (i) )         
                  ∠COA =  ∠BOD ( v. opp. angles ) 
                  OA = OD ( given ) 
by AAS ,  ΔCAO congruent to  Δ BDO 
      .'. OB = OC ( c.p.c.t.)
by the property of isosceles triangle two sides are equal and in the given  figure we proved that OB = OC therefore Δ OCB is isosceles triangle ....
                  ( proved ) 



  
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2015-04-01T12:56:43+05:30
GIVEN :- OA = OD and Angle 1 = angle 2
To Prove :- Triangle OCB is an isosceles triangle
Proof:-
As Angle 1 = angle 2
Then, 180 - angle 1 = 180 - angle 2
therefore, angle OAC = angle ODB                    - (i)

Now in Triangle OAC and ODB, we have:-
angle OAC = angle ODB                              (from i)
OA = OD                                                    (Given)
angle COA = angle BOD                              (Vertically Opposite angles)

So, Triangle OAC congruent to traingle ODB by ASA criterion of congruency
So, By C.P.C.T we get:-
OC = OB

Therefore, Triangle OCB is an isosceles triangle.
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