UPSC 2015 CDS II Math Paper

O is equidistant from the three sides of ΔABC. Therefore, O is the incentre of ΔABC.
The incentre is the point of intersection of the bisectors of the angles of the triangle.
∴∠OAC=

∠A

2

,∠OAB=

∠C

2

, and ∠OBA=

∠B

2

In ΔABC,
∠A+∠B+∠C=180°
⇒

∠A

2

+

∠B

2

+

∠C

2

=90°
⇒∠OAC+∠OBA+∠OCB=90°
Hence, statement 1 is correct.Remember:
The angle subtended by the arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
Now,∠BOC=2∠BAC if O is the circumcentre of ABC. But, O is the incentre of ΔABC. This is possible in case of an equilateral triangle only.
So,∠BOC=2∠BACdoes notalways hold.
Hence, statement 2 is not correct
Remember: Any point on the bisector of the angle between two intersecting lines is equidistant from the given lines.
Thus, the perpendiculars drawn from any point on OA to AB and AC are always equal. Hence, statement 3 is correct.