Question

solve it plz...if u can ​

Answer 1

Answer:

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.

hope it helps you....

Answer 2

angle 1 = angle 2

180° - angle 1 = 180° - angle 2

... (subtracting both sides from 180°)

angle OAC = angle ODB ... (linear pairs)

In ∆AOC and ∆DOB

angle OAC = angle ODB ... (from above)

OA = OD ... (given)

angle AOC = angle DOB

... (vertically opposite angles)

∆AOC $\frac{ ~ }{ = }$ ∆DOB

... (by ASA congruency test)

hence,

OC = OB ... (c. p. c. t.)

Hence, ∆BOC is isosceles

... Hence Proved!