Theorems on Concurrence in a Triangle

In geometry, three or more lines are called concurrent when they pass through the same point. A triangle has four important concurrence results involving its perpendicular bisectors, altitudes, internal angle bisectors and medians.

🔎 Important results at a glance

👉 The three perpendicular bisectors of the sides of a triangle are concurrent. Their common point is the circumcenter.

👉 The three altitudes of a triangle are concurrent. Their common point is the orthocenter.

👉 The three internal angle bisectors of a triangle are concurrent. Their common point is the incenter.

👉 The three medians of a triangle are concurrent. Their common point is the centroid.

⭐ The three perpendicular bisectors of the sides of a triangle are concurrent.

D, E and F are the midpoints of AB, BC and CA respectively in △ABC

Given:	In △ABC, D, E and F are the midpoints of AB, BC and CA respectively. The perpendiculars to AB at D and to BC at E meet at O.
To prove:	The perpendicular bisectors of AB, BC and CA are concurrent. It is enough to prove OF ⟂ AC, showing that O also lies on the perpendicular bisector of AC.
Construction:	Join O to A, B and C.
Proof:	In △AOD and △BOD,
	AD = BD [D is the midpoint of AB]
	∠ADO = ∠BDO = 90° [OD ⟂ AB]
	OD is common.
Therefore,	△AOD ≅ △BOD [SAS]
Hence,	OA = OB [Corresponding sides of congruent triangles] ...(i)
Similarly,	△BOE ≅ △COE [SAS]
Hence,	OB = OC ...(ii)
From (i) and (ii),	OA = OC.
Now,	In △AFO and △CFO,
	OA = OC [Proved above]
	AF = CF [F is the midpoint of AC]
	OF is common.
Therefore,	△AFO ≅ △CFO [SSS]
Hence,	∠AFO = ∠CFO [Corresponding angles of congruent triangles]
But,	∠AFO and ∠CFO form a linear pair on straight line AC. Since they are equal, each is 90°.
Therefore,	OF ⟂ AC.
Hence,	The three perpendicular bisectors of the sides of △ABC are concurrent. [Proved]

The point of concurrence is called the circumcenter. Since it lies on the perpendicular bisectors of all three sides, it is equidistant from the three vertices:

OA = OB = OC.

⭐ The three altitudes of a triangle are concurrent.

AD, BE and CF are perpendiculars from A, B and C to the opposite sides of △ABC

Given:	In △ABC, AD, BE and CF are drawn from A, B and C perpendicular to the opposite sides BC, CA and AB respectively.
To prove:	AD, BE and CF are concurrent.
Construction:	Through A, B and C draw lines parallel to BC, CA and AB respectively. Let these lines meet one another at P, Q and R, forming △PQR.
Proof:	By construction, APBC, ABCR and ABQC are parallelograms.
From parallelogram APBC,	AP = BC.
From parallelogram ABCR,	AR = BC.
Therefore,	AP = AR. Hence A is the midpoint of PR.
Similarly,	B is the midpoint of PQ and C is the midpoint of QR.
Again,	PR ∥ BC [By construction]
and	AD ⟂ BC [Given]
Therefore,	AD ⟂ PR [Since PR ∥ BC]
Thus,	AD is the perpendicular bisector of PR in △PQR.
Similarly,	BE is the perpendicular bisector of PQ and CF is the perpendicular bisector of QR.
But,	The three perpendicular bisectors of the sides of a triangle are concurrent.
Therefore,	AD, BE and CF are concurrent.
Hence,	The three altitudes of △ABC are concurrent. [Proved]

The common point of the three altitudes is called the orthocenter.

⭐ The three internal angle bisectors of a triangle are concurrent.

Internal angle bisectors of ∠A, ∠B and ∠C in △ABC

Given:	In △ABC, the internal bisectors of ∠B and ∠C meet at O. Join A and O.
To prove:	The internal bisectors of ∠A, ∠B and ∠C are concurrent. In other words, prove that AO bisects ∠A.
Construction:	From O draw OP, OQ and OR perpendicular to AB, BC and AC respectively.
Proof:	In △BOQ and △BOP,
	∠OBQ = ∠OBP [BO bisects ∠B]
	∠OQB = ∠OPB = 90° [OQ ⟂ BC and OP ⟂ AB]
	BO is common.
Therefore,	△BOQ ≅ △BOP [AAS]
Hence,	OQ = OP [Corresponding sides of congruent triangles] ...(i)
Similarly,	△COQ ≅ △COR [AAS]
Hence,	OQ = OR ...(ii)
From (i) and (ii),	OP = OR ...(iii)
Now,	Consider right triangles △APO and △ARO.
	∠OPA = ∠ORA = 90°.
	AO is the common hypotenuse.
	OP = OR [From (iii)]
Therefore,	△APO ≅ △ARO [RHS]
Hence,	∠PAO = ∠RAO [Corresponding angles of congruent triangles]
Therefore,	AO bisects ∠A.
Hence,	The three internal angle bisectors of △ABC are concurrent. [Proved]

The common point is called the incenter. It is equidistant from all three sides of the triangle.

⭐ The three medians of a triangle are concurrent.

The medians AD, BE and CF of △ABC meet at G

Given:	In △ABC, medians BE and CF meet at G. Join A to G and produce AG to meet BC at D.
To prove:	AD is the third median. That is, prove that D is the midpoint of BC. Then all three medians will be concurrent.
Construction:	Produce AD to H so that AG = GH. Join B to H and C to H.
Proof:	In △ABH, F is the midpoint of AB [CF is a median].
and	G is the midpoint of AH [AG = GH].
Therefore,	FG ∥ BH [Midpoint theorem]
But,	F, G and C are collinear [CF passes through G].
Therefore,	GC ∥ BH.
Again,	In △ACH, E is the midpoint of AC [BE is a median].
and	G is the midpoint of AH.
Therefore,	GE ∥ HC [Midpoint theorem]
But,	B, G and E are collinear [BE passes through G].
Therefore,	BG ∥ HC.
Now,	In quadrilateral BGCH, GC ∥ BH and BG ∥ HC.
Therefore,	BGCH is a parallelogram.
Hence,	The diagonals BC and GH of parallelogram BGCH bisect each other.
Therefore,	D is the midpoint of BC.
Hence,	AD is the third median of △ABC.
Therefore,	The three medians of a triangle are concurrent. [Proved]

The common point of the medians is called the centroid. The centroid divides every median in the ratio 2:1 from the vertex.

🧩 Solved Examples

Example 1: Circumcenter and equal distances

The perpendicular bisectors of the sides of △ABC meet at O. If OA = 6 cm, find OB and OC.

Solution: O is the circumcenter, so it is equidistant from all three vertices.

OA = OB = OC

Therefore, OB = 6 cm and OC = 6 cm.

Example 2: Distance of the incenter from the sides

The internal angle bisectors of a triangle meet at I. The perpendicular distance from I to one side is 4 cm. Find the perpendicular distance from I to each of the other two sides.

Solution: I is the incenter, and the incenter is equidistant from all three sides.

Therefore, each of the other two perpendicular distances is 4 cm.

Example 3: Centroid and a median

In △ABC, G is the centroid and AD is a median. If AD = 12 cm, find AG and GD.

Solution: The centroid divides a median in the ratio 2:1 from the vertex.

AG = 2/3 × 12 = 8 cm GD = 1/3 × 12 = 4 cm

✍️ Practice Problems

The perpendicular bisectors of the sides of △PQR meet at O. If OP = 7 cm, find OQ and OR.
Name the point where the three altitudes of a triangle meet.
The internal angle bisectors of △ABC meet at I. Perpendiculars from I to AB, BC and CA have lengths 5 cm, x cm and y cm respectively. Find x and y.
In △XYZ, G is the centroid and XM is a median of length 15 cm. Find XG and GM.
Which triangle centre is always inside a triangle: circumcenter, orthocenter, incenter or all of them?
In an equilateral triangle, what can you say about the circumcenter, orthocenter, incenter and centroid?

✅ Answers

1. OQ = OR = 7 cm.

2. Orthocenter.

3. x = 5 cm and y = 5 cm.

4. XG = 10 cm and GM = 5 cm.

5. The incenter is always inside the triangle.

6. All four centres coincide at the same point.

📚 Important Points to Remember

👉 Three or more lines are called concurrent when they pass through the same point.

👉 The three perpendicular bisectors of a triangle meet at the circumcenter. The circumcenter is equidistant from the three vertices.

👉 The three altitudes meet at the orthocenter.

👉 The three internal angle bisectors meet at the incenter. The incenter is equidistant from the three sides.

👉 The three medians meet at the centroid.

👉 The centroid divides each median in the ratio 2:1 from the vertex.

👉 A perpendicular bisector of a side and an angle bisector are different. A perpendicular bisector is related to a side, while an angle bisector divides an angle into two equal angles.

👉 These four concurrence theorems are true for every triangle, but the four centres are generally different points.

👉 In an equilateral triangle, the circumcenter, orthocenter, incenter and centroid all coincide.

👉 Proofs of concurrence often use congruent triangles, the midpoint theorem, parallel lines and properties of parallelograms.