in the figure,if OA=OD AND ANGLE1=ANGLE2.Prove that triangle OCB is an isosceles triangle.
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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 )
∠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 )
Answered by
61
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.
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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