CBSE Class 9 Mathematics Loci and Concurrent Lines in Triangles Notes Set 05

Locus

A locus is the set of all points that satisfy a given geometrical condition. Two things must hold: every point satisfying the condition lies on the locus, and every point of the locus satisfies the condition.
1. The locus of points at a fixed distance r from a fixed point O is the circle with centre O and radius r.
2. The locus of points equidistant from two fixed points A and B is the perpendicular bisector of AB.
3. The locus of points equidistant from two intersecting lines is the pair of bisectors of the angles between them.
4. The locus of points equidistant from the arms of an angle (inside it) is its bisector.

Centres of a Triangle

Lines associated with a triangle meet in single points, called points of concurrency.
1. Incentre I: the three angle bisectors meet at I. It is equidistant from the three sides, so it is the centre of the incircle and the distance is the inradius.
2. Circumcentre O: the three perpendicular bisectors of the sides meet at O. It is equidistant from the three vertices; this distance is the circumradius.
3. Centroid G: the three medians meet at G, which divides each median in the ratio 2 : 1 counted from the vertex.
4. Orthocentre H: the three altitudes meet at H.

Position of the Circumcentre and Orthocentre

1. Acute-angled triangle: circumcentre and orthocentre both lie inside the triangle.
2. Right-angled triangle: the circumcentre is the midpoint of the hypotenuse and the orthocentre is the vertex of the right angle.
3. Obtuse-angled triangle: both lie outside the triangle.
4. The centroid and the incentre always lie inside the triangle.
5. In an equilateral triangle all four centres coincide. In an isosceles triangle they all lie on the axis of symmetry.

Key Proofs in Brief

1. Medians meet at one point: let two medians BE and CF meet at G and extend AG to P with GP = AG. Then BGCP is a parallelogram because FG joins midpoints in triangle ABP and GE in triangle ACP. Its diagonals bisect each other, so AP passes through the midpoint D of BC. Hence the third median passes through G, and G divides each median in the ratio 2 : 1.
2. Angle bisectors meet at one point: a point on the bisector of angle B is equidistant from BA and BC; a point on the bisector of angle C is equidistant from CB and CA. Their meeting point is therefore equidistant from all three sides, so it lies on the bisector of angle A as well.
3. Perpendicular bisectors meet at one point: a point on the bisector of AB is equidistant from A and B, and a point on the bisector of BC is equidistant from B and C. The meeting point is equidistant from A and C, so it also lies on the bisector of AC.

Constructions of Triangles

1. Given base BC, base angle B and the sum AB + AC: draw BC and a ray BX at the given angle, cut BD equal to the given sum, join DC and take A where the perpendicular bisector of DC meets BD.
2. Given base, base angle and the difference of the other two sides: cut off the difference on the ray (or its extension) and use the perpendicular bisector in the same way.
3. Given the perimeter and the two base angles: draw a line segment equal to the perimeter, bisect the angles equal to half of each base angle at its ends to get the vertex, then use the perpendicular bisectors of the two joining lines to locate the other two vertices on the segment.