JEE Mathematics Concurrent Lines in a Triangle Notes Set 02

Important Notes and Formulas

Point of Concurrency

When three or more lines pass through one common point, the lines are said to be concurrent and the common point is their point of concurrency. A triangle has four important sets of concurrent lines, and each gives a special point.

Incentre

The internal bisectors of the three angles of a triangle meet at a single point I, called the incentre.
I is at the same distance from all three sides. The circle with centre I and this distance as radius touches all three sides from inside; it is the incircle and its radius is the inradius.
The incentre always lies inside the triangle.

Circumcentre

The perpendicular bisectors of the three sides meet at a single point O, called the circumcentre.
O is at the same distance from the three vertices. This distance is the circumradius, and the circle through all three vertices is the circumcircle.
Position of O: inside the triangle for an acute-angled triangle, at the midpoint of the hypotenuse for a right-angled triangle, and outside the triangle for an obtuse-angled triangle.

Centroid

A median joins a vertex to the midpoint of the opposite side. The three medians meet at a point G, called the centroid.
G divides every median in the ratio \( 2:1 \), the longer part being towards the vertex. The centroid always lies inside the triangle.

Orthocentre

An altitude is the perpendicular from a vertex to the opposite side. The three altitudes meet at a point H, called the orthocentre.
H is inside an acute triangle, at the right-angle vertex of a right triangle, and outside an obtuse triangle.

Medians Are Concurrent: Proof Idea

Let the medians BE and CF meet at G. Extend AG to a point P with GP = AG, and join BP and CP.
In triangle ABP the segment FG joins the midpoints of AB and AP, so FG is parallel to BP. In the same way GE is parallel to CP. Hence BGCP is a parallelogram.
The diagonals of a parallelogram bisect each other, so they meet at the midpoint D of BC. Thus AD is also a median and passes through G.
Finally \( AG = GP = 2\,GD \), so \( AG:GD = 2:1 \). The same argument for the other medians gives \( BG:GE = CG:GF = 2:1 \).

Locus

A locus is the set of all points that satisfy a given condition. Two things must be true: every point satisfying the condition lies on the locus, and every point of the locus satisfies the condition.
The locus of points at a fixed distance \( r \) from a fixed point O is the circle with centre O and radius \( r \).
The locus of points equidistant from two parallel lines is a third line parallel to both, midway between them.

Locus of a Point Equidistant from Two Points

Theorem: the locus of a point equidistant from two given points A and B is the perpendicular bisector of the segment AB.
Why: if M is the midpoint of AB and P satisfies PA = PB, then triangles AMP and BMP have three equal sides, so they are congruent. This gives \( \angle PMA = \angle PMB = 90^\circ \), so P lies on the perpendicular bisector of AB.
Conversely, every point Q of the perpendicular bisector satisfies QA = QB, because triangles AMQ and BMQ are congruent (two sides and the included right angle).

Locus of a Point Equidistant from Two Intersecting Lines

Theorem: the locus of a point equidistant from two intersecting lines is the pair of lines that bisect the angles formed by them.
Why: if P is inside one of the angles and its perpendicular distances PM and PN to the two lines are equal, the right triangles OPM and OPN are congruent (right angle, common hypotenuse OP, PM = PN). So P lies on the bisector of that angle. Points in the other three angles are handled in the same way.
Conversely, a point on either bisector has equal perpendicular distances to the two lines.

Solved Examples

Example 1. In triangle ABC the bisectors of angles B and C meet at I. Show that AI bisects angle A.
Answer:
Draw perpendiculars IP, IQ, IR from I to BC, CA, AB.
I lies on the bisector of angle B, so triangles BPI and BRI are congruent (two angles and the common side BI). Hence IP = IR.
In the same way triangles CPI and CQI are congruent, so IP = IQ.
Therefore IQ = IR. The right triangles AQI and ARI have a common hypotenuse AI and IQ = IR, so they are congruent and \( \angle IAQ = \angle IAR \).
Hence AI bisects angle A, which also shows that all three internal bisectors pass through I.

Example 2. Triangles PBC, QBC and RBC are isosceles triangles on the same base BC (with PB = PC, QB = QC, RB = RC). Show that P, Q and R are collinear.
Answer:
Since PB = PC, the point P is equidistant from B and C, so P lies on the perpendicular bisector of BC.
Similarly Q and R lie on the perpendicular bisector of BC.
Three points on one line are collinear, so P, Q and R are collinear.

Example 3. PBC and QBC are isosceles triangles on the same side of the same base BC (PB = PC and QB = QC). Show that PQ bisects BC at right angles.
Answer:
P and Q are both equidistant from B and C, so both lie on the perpendicular bisector of BC.
The line through P and Q is therefore that perpendicular bisector.
So PQ bisects BC and is perpendicular to it.

Example 4. The bisectors of angles B and C of a quadrilateral ABCD meet at P. Show that P is equidistant from the opposite sides AB and CD.
Answer:
Draw PM perpendicular to AB, PN perpendicular to CD and PL perpendicular to BC.
P lies on the bisector of angle B, so it is equidistant from AB and BC: PM = PL.
P lies on the bisector of angle C, so it is equidistant from BC and CD: PN = PL.
Hence PM = PN, so P is equidistant from AB and CD.

Example 5. In a right-angled triangle with hypotenuse 10 cm, find the circumradius.
Answer:
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse.
So the circumradius is half the hypotenuse, \( \dfrac{10}{2} = 5 \) cm.