if o is the circumcentre ofa triangle abc and od is perpendicular on bc then angle bod must be equal to
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i think in this triangle,BOD=COD
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Two triangles abc and obc are formed inside the circle. Where o is the circumcentre and abc are vertices of triangle.
(Angle boc)= 2(angle bac) [Theorem: Angle formed by the arc of the circle with the centre is double the angle which it subtends on the remaining part of circle]
As o is the circumcentre implies od is the perpendicular bisector of bc. Hence
angle boc = angle bod + angle cod
As bd=dc, angle bod=angle cod (If sides are equal their opposite angles are also equal)
Thus, 2(angle bod) = angle boc
But angle boc = 2(angle bac), therefore,
Angle bod= Angle bac.
(Angle boc)= 2(angle bac) [Theorem: Angle formed by the arc of the circle with the centre is double the angle which it subtends on the remaining part of circle]
As o is the circumcentre implies od is the perpendicular bisector of bc. Hence
angle boc = angle bod + angle cod
As bd=dc, angle bod=angle cod (If sides are equal their opposite angles are also equal)
Thus, 2(angle bod) = angle boc
But angle boc = 2(angle bac), therefore,
Angle bod= Angle bac.
man:
M not fully sure about the answer being correct, check it for urself and kindly tell me if i m wrong somewhere.
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