JEE Mathematics Concurrent Lines in a Triangle Notes Set 01

Special Points of a Triangle

Two Basic Locus Results

1. The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.
2. The locus of points equidistant from the two arms of an angle (inside the angle) is the bisector of that angle.
3. Consequence: a circle through two given points has its centre on the perpendicular bisector of the segment joining them.
4. Consequence: three non-collinear points are equidistant from exactly one point, the circumcentre.

Why the Perpendicular Bisectors Meet

Let the bisectors of \(AB\) and \(AC\) meet at \(O\). Then \(OA=OB\) and \(OA=OC\), so \(OB=OC\), which places \(O\) on the bisector of \(BC\) as well. The three lines are therefore concurrent.

Why the Altitudes Meet

Draw through each vertex a line parallel to the opposite side. These lines form a larger triangle in which the vertices of the original triangle are the mid-points of the sides, because each pair of parallels forms a parallelogram. The altitudes of the original triangle are then the perpendicular bisectors of the sides of the larger triangle, and these are concurrent.

Medians

1. The centroid \(G\) satisfies \(AG:GD=2:1\) for the median \(AD\).
2. The sum of any two medians is greater than the third median.
3. If two medians are equal, the triangle is isosceles, with the two sides from which those medians start being equal.

Isosceles Triangle

If \(AB=AC\) and \(D\) is the mid-point of \(BC\), the line \(AD\) is at once the median, the altitude, the angle bisector and the perpendicular bisector of \(BC\). The circumcentre, orthocentre, centroid and incentre all lie on it.

Orthocentric System

If \(P\) is the orthocentre of triangle \(ABC\), then \(A\) is the orthocentre of triangle \(PBC\), \(B\) of \(PCA\) and \(C\) of \(PAB\).