The circumcenter of a triangle is 0 prove that â obc+â bac = 90o
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inscribed angle BAC = a
=> angle BOC = 2a ( as angle subtended by chord BC at the centre of the circle is twice the inscribed angle.
< BOD = < COD = a ( as tri ODB is congruent to tri ODC)
In right triangle ODB,
< ODB = 90° ( as OF is perpendicular to BC)
< BOD = a
So, < OBD = 180- (90+a) = 90 - a = < OBC
=> angle BOC = 2a ( as angle subtended by chord BC at the centre of the circle is twice the inscribed angle.
< BOD = < COD = a ( as tri ODB is congruent to tri ODC)
In right triangle ODB,
< ODB = 90° ( as OF is perpendicular to BC)
< BOD = a
So, < OBD = 180- (90+a) = 90 - a = < OBC
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