The circumcentre of the triangle ABC is O . Prove that `angle OBC+angle BAC=90^@`.

Explanation

ABC is a triangle and O is its circumcentre.
Draw `OdbotBC`. Join OB and OC.
In right `Delta OBD` and right `Delta OCD`, we have
hypotenuse OB=hyotenuse OC radii of the same circle)
`OD=OD` (common side)
`therefore Delta OBDcongDeltaOCD` (by R.H.S. congruency rule)
`therefore angle1=angle2` and `angle3 =angle4`(c.p.c.t)
Now, `angleBOC =2angle1` and `angle BOC =2angleA`
`therefore2angle1=2anglerArrangle1=angleA`
`thereforeangleA=angle2`
`(becauseangle1 =angle 2 )`........(1)
`rArrangleA+ angle3=90^@ " "(because angle2+angle4=90^@ and angle4=angle3)`
`rArrangle OBC +angle A=90^@`.