The point 0 is equidistant from the three sides of a triangle ABC Consider the following statements ...........

Quantitative Aptitude (CDS) Coordinate Geometry

Download Solution PDF

The point 0 is equidistant from the three sides of a triangle ABC. Consider the following statements:

1 ∠OAC + ∠OCB + ∠OBA = 90°

2. ∠BOC = 2∠BAC

3. The perpendiculars drawn from any point on OA to AB and AC are always equal

Which of the above statements are correct?

1 and 2 only

2 and 3 only

1 and 3 only

1, 2 and 3

छोटे Courses बड़े Results

Master Coordinate Geometry

Prepare through Micro-courses

Try Now

Explanation:

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