Theorems On Concurrence

Let us consider an acute angle triangle PQR.

OA, OB and OC perpendicularly bisects the sides.

Joined OP, OQ and OR!

In Δ AOQ and Δ AOR

AO = AO (Common side)

AQ = AR(A is the midpoint of QR)

∠OAQ = ∠OAR (OA is the perpendicular bisector)

Hence Δ QOA and Δ ROA are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OR (Corresponding parts of Congruent Triangles)

In Δ QOC and Δ POC

CO = CO (Common side)

CQ = CP (C is the midpoint of QP)

∠OCQ = ∠OCP (OC is the perpendicular bisector)

Hence Δ QOC and Δ POC are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OP (Corresponding parts of Congruent Triangles)

Hence OP = OQ = OR

So it is proved that the perpendicular bisectors are concurrent

The circumcentre lies inside in an acute angled triangle

Let us consider an obtuse angled triangle PQR.

OA, OB and OC perpendicularly bisect the sides.

Joined OP, OQ and OR!

In Δ AOQ and Δ AOR

AO = AO (Common side)

AQ = AR (A is the midpoint of QR)

∠OAQ = ∠OAR (OA is the perpendicular bisector)

Hence Δ QOA and Δ ROA are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OR (Corresponding parts of Congruent Triangles)

In Δ QOC and Δ POC

CO = CO (Common side)

CQ = CP(C is the midpoint of QP)

∠OCQ = ∠OCP (OC is the perpendicular bisector)

Hence Δ QOC and Δ POC are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OP (Corresponding parts of Congruent Triangles)

Hence OP = OQ = OR

So it is proved that the perpendicular bisectors are concurrent

The circum centre lies outside in an obtuse angled triangle

PO is perpendicular to AB and OQ is perpendicular to BC

Since Δ ABC is right angled at B so POQB forms a rectangle

Hence we can say that

PO = BQ (Opposite sides of rectangle)

BP = QO(Opposite sides of a rectangle)

In Δ AOP and Δ QOC

∠APO = ∠OQC (OP and OQ are perpendiculars)

∠OAP = ∠COQ (Corresponding angles)

PO = BQ = CQ (Q is midpoint of BC)

So Δ AOP and Δ COQ are congruent by A.A.S. axiom of congruency

Hence AO = OC

So O is the midpoint of AC

Hence the perpendiculars are concurrent

The circum centre lies on the midpoint of the hypotenuse of the triangle.

The Δ PQR is right angled at Q

Let PQ = 8 cm

QR = 6 cm

PR = 10 cm

Squaring and adding the smaller two sides of the triangle we get

PQ² + QR² = 6² + 8²

PQ² + QR² = 36 + 64

⇒ PQ² + QR² = 100

⇒ PQ² + QR² = 10²

⇒ PQ² + QR² = PR²

So we see the sum of squares of the two smaller sides is equal to the square of the larger side.

Hence the triangle is a right angled triangle.

The circum radius of a right angled triangle is always half of the length of the hypotenuse.

Circum radius

In a right angled triangle the length of hypotenuse is always twice the length of its circum radius

The Δ PQR is right angled at Q

PR is the hypotenuse

OP = OR is the circum radius

Length of circum radius = 10 cm

⇒ Length of hypotenuse = 2×10 = 20 cm

_{Let∠ABI =∠CBI = x}

_{And∠ACI =∠BCI = y}

_{In Δ ABC, using angle sum property,}

_{∠A + 2x + 2y = 180° ….(1)}

_{In Δ BIC, using angle sum property,}

_{∠I + x + y = 180° ….(2)}

_{Divide (1) by 2}

_{We get,}

_{Subtract (3) from (2)}

_Or

Given:

As we know,

If O is circumcenter of Δ ABC then:

∠BOC = 2 ∠BAC (Application 1)

⇒ ∠BAC = 40°

Option (A) is correct.

Given:

As we know,

If O is orthocenter Δ ABC then sum of ∠BOC and ∠BAC is180°:

i.e. ∠BOC + ∠BAC =180°:

∠BOC =180° - ∠BAC

⇒ ∠BOC = 180° - 40°

⇒ ∠BAC = 140°

Option (B) is correct.

Given:

As we know,

If O is incenter of ΔABC then:

⇒ ∠BOC = 90° + 20°

⇒ ∠BOC = 110°

Option (B) is correct.

Given: area of GBC is 12 sq cm

As we know if three medians AD, BE and CF of a ΔABC intersect one another at the point G (centroid) then,

Area of ΔABC = 3 × Δ BGC

⇒ Area of ΔABC = 3 × 12

⇒ Area of ΔABC = 36 sq. cm

Option (C) ic correct.

Given: length of circumradius of a right angled triangle = 5 cm

As we know,

length of circumradius of a right angled triangle = 1/2 length of hypotenuse

⇒ BD = 1/2 AC

⇒ AC = 2 BD

⇒ AC = 2 (5) = 10cm

Option (B) ic correct.

In Δ BGD and Δ CGE,

∠BGD = ∠CGE (vertically opposite angles) …(1)

BE = DC (medians are equal)

Since, centroid divides the median in ratio 2:1

So, and

⇒ BG = CG (as BE = DC) …(2)

And and

⇒ GE = DG (BE = DC) …..(3)

Hence, by SAS congruency, Δ BGD and Δ CGE are congruent.

By CPCT, BD = EC

2× BD = 2 × EC

⇒ AB = AC

Hence, the triangle is isosceles.

We know,

Circumcenter of a triangle: The intersection of perpendicular bisectors of the triangles.

Incenter: The intersection of angle bisectors of the triangle.

Centroid: The intersection of the medians of the triangle.

Orthocentr: The intersection of the altitudes of the triangle

In order to prove that, In an equilateral triangle, circumcenter, incentre, centroid and orthocenter coincide, it is sufficient to prove that for any side median, altitude, perpendicular bisector and angle bisector of angle opposite to that side is common.

Let us consider an equilateral ΔABC, such that AD is a median to side BC.

To Prove:

(i) AD ⊥ BC [AD is Altitude and perpendicular bisector of BC]

(ii) ∠BAD= ∠CAD [AD is angle bisector of ∠A]

Proof:

In ΔABD and ΔACD

AB = AC [Sides of equilateral triangle]

AD = AD [Common side]

BD = CD [As, AD is median to BC]

⇒ ΔABD ≅ ΔACD [By SSS property of congruent triangles]

Now, As corresponding parts of congruent triangles are equal [CPCT], we have

∠ADB = ∠ADC

Also,

∠ADB + ∠ADC = 180° [Linear pair]

⇒ ∠ADB + ∠ADB = 180°

⇒ ∠ADB = ∠ADC = 90°

⇒ AD ⊥ BC

Since, AD ⊥ BC, and BD = BC

∴ AD is perpendicular bisector of BC and as well as altitude from A to BC.

Now,

∠BAD = ∠CAD [By CPCT]

AD is angle bisector of ∠A.

∴ AD is median, perpendicular bisector, altitude and angle- bisector.

We can prove this result for every median, and since lines are same, their intersection will also be same.

Hence Proved!

Given: A triangle ABC, AD BE and CF are medians, let us join DEF to make a triangle and label the points where medians of ΔABC intersects the sides of ΔDEF. Let O be the intersection of medians of ΔABC i.e. O is centroid of ΔABC

To Prove: O is centroid of ΔDEF

Proof:

As, CF and BE are medians to sides AB and AC respectively,

F is mid-point of AB and E is mid-point of AC

By mid-point theorem [The line segment connecting the midpoints of two sides of a triangle is parallel to the third side]

EF || BC

⇒ FH || BD and HE || DC

In ΔABD and ΔAFH

∠FAH = ∠BAD [Common]

∠AFH = ∠ABD [Alternate angles]

ΔABD ∼ ΔAFH [By AA similarity criterion]

……[1] [Similar triangles have equal ratio of Corresponding sides]

In ΔADC and ΔAHE

∠HAE = ∠DAC [Common]

∠AHE = ∠ADC [Alternate angles]

ΔADC ∼ ΔAHE [By AA similarity criterion]

……[2] [Similar triangles have equal ratio of Corresponding sides]

From [1] and [2], we have

Here, BD = DC ∵ AD is median

⇒ FH = HE

⇒ DH is a median from D to side FH

Similarly, we can prove that EI and FG are medians, and as intersection of FH, EI and FG is O.

⇒ O is centroid for ΔDEF

⇒ ΔDEF and ΔABC have same centroid.

Hence, Proved!

Given: AD, BE and FC are medians.

Extend AD to H such that (AG = HG) then join BH and CH.

In Δ ABH,

F is mid-point and G is centroid, so use mid-point theorem,

So, FG || BH

Similarly, GC || BH ….(1)

And BG || HC …..(2)

From (1) and (2),

We get,

BGCH is a parallelogram,

So, BH = GC …..(3) (opp. Sides of a parallelogram are equal)

In Δ BGH,

BG + GH > BH (sum of two sides greater than third side)

⇒ BG + AG > GC

Similarly, BE + CF > AD and AD + CF > BE

Hence proved.

(i)

As we know, if G is centroid, and centroid divides the median in 2:1

Then BE = BG and FC = CG

In

Δ BGC,

BG + GC > BC (sum of two sides greater than 3^rd side)

⇒

⇒ 2BE + 2FC > 3BC

Or

3BC < 2BE + 2FC ….(1)

Similarly,

3CA < 2CF + 2AD ….(2)

And 3AB < 2AD + 2BE ….(3)

Add (1), (2) and (3)

We get,

3 BC + 3CA + 3AB < 2BE + 2FC + 2CF + 2AD + 2AD + 2BE

⇒ 3 (AB + BC + CA) = 4 BE + 4 FC + 4 AD

Or 4 (AD + BE + CF) > 3 (AB + BC + CA)

Hence proved.

(ii) In ΔACD

AC + CD > AD (sum of two sides greater than 3^rd side)

Since, AD is median

……[1]

Similarly,

In ΔABE

……[2]

In ΔBFC

……[3]

Adding [1], [2] and [3], we get

Hence, Proved!

Given: area of ABC is 36 sq cm

(i) As we know if three medians AD, BE and CF of a ΔABC intersect one another at the point G (centroid) then,

Area of ΔABC = 3 × Δ AGB

⇒ Area of ΔAGB =

⇒ Area of ΔAGB = 12 sq. cm

(ii) As we know if three medians AD, BE and CF of a ΔABC intersect one another at the point G (centroid) then,

Area of ΔABC = 6 × Δ GEC

⇒ Area of ΔGEC =

⇒ Area of ΔGEC = 6 sq. cm

(ii) As we know if three medians AD, BE and CF of a ΔABC intersect one another at the point G (centroid) then,

Area of ΔABC = 3 × Δ BDGF

⇒ Area of ΔBDGF =

⇒ Area of ΔBDGF = 12 sq. cm

Given: AD = BC

As we know, if G is centroid,

Then AD = AG

⇒ AG = BC

⇒ if AG = 2 then GD = 1 ….(1)

Also, BC = 2 (BG = AG)

BD = DC = 1 (median) ….(2)

From (1) and (2)

BD = DC = GD = 1

⇒ ∠GDB = 90° and ∠GDC = 90°

Because, BD = GD

⇒ ∠GBD = ∠DGB = x (let) (isosceles triangle)

Hence,

x + x + 90° = 180°

2x = 90°

⇒ x = 45°

Similarly,

Because, DC = DG

⇒ ∠GCD = ∠DGC = y (let) (isosceles triangle)

Hence,

y + y + 90° = 180°

2y = 90°

⇒ y = 45°

And x + y = 45° + 45°

⇒ ∠BGC = 90°

i.e. angle between median is 90° .

Given: P and Q are the mid points of sides BC and CD of a parallelogram ABCD respectively; the diagonals AP and AQ cut BD at the points K and L

To Prove: BK = KL = LD.

Given: lengths of sides of triangle are 6 cm, 8 cm, and 10 cm.

Let AC = 10cm, BC = 8cm and AB = 6cm

To be a right angle triangle,

(AC)² = (AB)² + (BC)²

(10)² = (8)² + (6)²

100 = 64 + 36

100 = 100

Hence, LHS = RHS

So, it is a right angle triangle.

In a right angle triangle circumcentre lies on hypotenuse.

Given: length of side = a = 3√3 m

As we know,

Centroid is equidistant from the corners of an equilateral triangle.

Hence, AG is actually the radius of circumcircle to be inscribed around triangle.

As we know, in an equilateral triangle,

Circumradius of an equilateral triangle (r) =

⇒

⇒ r = 3

Then, length of AG = 3m

The point which is equidistant from all sides of a triangle is called incenter.

Hence, there is only one point equidistant from sides of a triangle.

Given: Pedal triangle is a triangle formed by joining the mid-point of a triangle’s sides.

As we know, in an equilateral triangle, triangles formed after joining the mid points is also an equilateral triangle.

Also, the triangle formed joining the vertx and one of the vertex is also an equilateral triangle.

As we know, the measure of an angle in an equilateral triangle is 60° .

Hence, in Δ DEF

∠FDE = 60°

And, in Δ ADE

∠ADE = 60°

So,

∠FDA = ∠FDE + ∠ADE = 60°

⇒ ∠FDA = 60° + 60°

⇒ ∠FDA = 120°

Given: Median AD = BC ….(1)

From (1) we can say it is a right angle triangle as well,

AB = √2 cm

AC = √2 cm (isosceles triangle)

In a right angle triangle,

(BC)² = (AB)² + (AC)²

⇒ (BC)² = (√2)² + (√2)²

⇒ (BC)²= 2 + 2

⇒ (BC)²= 4

⇒ BC = 2cm

As we know,

length of circumradius of a right angled triangle = 1/2 length of hypotenuse

⇒ AD = 1/2 BC

⇒ AD = 1/2 (2)

⇒ AD = 1cm