Question

In the given figure OEI BC and OD LAB such that OE=OD.
Is AOEB AODB? Give reasons for your answer.​

Answer 1

We know that BD=OD.

We know that BD=OD.But OD=OB= Radius of the circle.

We know that BD=OD.But OD=OB= Radius of the circle.Therefore

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OB

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°BDX= 30°

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°BDX= 30°We also know that,

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°BDX= 30°We also know that,BDC(BDX)= BAC (Angles lie on the same arc{BC} are equal in measure.

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°BDX= 30°We also know that,BDC(BDX)= BAC (Angles lie on the same arc{BC} are equal in measure.Therefore,

We know that BD=OD.But OD=OB= Radius of the circle.ThereforeBD=OD=OBBDO is equilateral triangle.Angle DBO= 60°Now let us take the intersecting point of CD and AB as X.In triangle BDX,BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)BXD+DBX+BDX=180°{Angle Sum Property}90°+60°+BDX= 180°BDX= 30°We also know that,BDC(BDX)= BAC (Angles lie on the same arc{BC} are equal in measure.Therefore,BAC=BDC(BDX)=30

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