WBBSE Solutions For Class 9 Maths Geometry Chapter 4 Theorems On Concurrence

Geometry Chapter 4 Theorems On Concurrence

⇔ Concurrent lines: If two or more different straight lines having a common point is said to be concurrent lines.

⇒ AB, CD, EF and GH are concurrent lines.

⇔ Circumcentre of a triangle: The point where the three perpendicular bisectors of sides of a triangle intersect is called the circumcentre of the triangle.

⇒ In ΔABC, the three perpendicular bisectors of AB, BC and CA meet at the point O; the point O is called the circumcentre and OA or OB or OC is the circumradius of ΔABC.

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⇒ [OA = OB = OC]

⇒ A circle is drawn passing through the points A, B and C is called circum circle.

⇔ Incentre of a triangle: The point where the three internal bisectors of angles of a triangle intersect is called the incentre of the triangle.

In ΔABC, the internal bisectors of angles ∠A, ∠B and ∠C intersect each other at the point O.

I drawn OD ⊥ BC;

A circle is drawn with centre O and length of equal radius of OD is said to be in circle of ΔABC.

The circle touch AC and AB at E and F respectively. Centre of the circle is called incentre.

⇔ Centroid of a triangle: The point where the three medians of a triangle intersect is said to be centroid of a triangle.

⇒ Three medians AD, BE and CF of a triangle ABC, intersect at the point G.

⇒ The point G is called the centroid of the ΔABC.

⇒ The centroid divides any median from the vertex in the ratio 2: 1.

⇔ Orthocentre of a triangle: The point where perpendiculars on the opposite sides from the three vertices of any triangle is called the orthocentre of the triangle.

⇒ In ΔABC, AD ⊥ BC, BE ⊥ CA and CF ⊥ AB;

⇒ AD, BE and CF meets at point O.

⇒ The point O is the orthocentre of ΔABC.

⇒ The triangle DEF obtained by joining the three points D, E, and F of ΔABC, is called a pedal triangle.

⇔ External centre: The point where the external bisectors of two angles and one internal bisector of an angle of a triangle intersect is called external centre.

⇒ In ΔABC, external bisectors of ∠ABC and ∠ACB and internal bisector of ∠BAC intersect at point O.

So O is the external centre.

OD, OE, OF are said to be external radius and the circle passes through the points D, E and F are called external circles of the triangle ABC.

Theorems:

1. The perpendicular bisectors of sides of triangle are concurrent.
2. The perpendiculars from vertices of triangle on the opposite sides are concurrent.
3. The internal bisectors of angles of triangle are concurrent.
4. The three medians of a triangle are concurrent.

Geometry Chapter 4 Theorems On Concurrence True Or False

Example 1. The sum of lengths of three medians of a triangle is greater than three-fourth of its perimeter.

Solution: In ΔABC, the medians AD, BE and CF intersects at G (centroid).

∴ \(\frac{A G}{G D}=\frac{2}{1}\)

[centroid of a triangle divides any median from the vertex in the ratio 2: 1]

⇒ \(\frac{G D}{A G}=\frac{1}{2}\)

⇒ \(\frac{G D}{A G}+1=\frac{1}{2}+1\)

⇒ \(\frac{G D+A G}{A G}=\frac{3}{2}\)

⇒ \(\text { i.e. } \frac{A D}{A G}=\frac{3}{2}\)

⇒ \(A G=\frac{2}{3} A D\)

Similarly, BG = \(\frac{2}{3}\)  BD and CG = \(\frac{2}{3}\) CF

In ΔABG, AG + BG > AB…….(1)

In ΔBCG, BG + CG > BC……..(2)

In, ΔACG, CG + AG > AC……….(3)

[The sum of lengths of two sides of a triangle is greater than the length of third side]

(1) + (2) + (3) we get

2(AG + BG + CG) > AB + BC + AC

2(\(\frac{2}{3}\) AD + \(\frac{2}{3}\) BE + \(\frac{2}{3}\) CF) > AB + BC + CA

⇒ \(\frac{4}{3}\)(AD + BE + CF) > AB + BC + CA

⇒ AD + BE + CF > \(\frac{3}{4}\) (AB + BC + CA)

So the statement is true.

Example 2. The orthocentre of a triangle is a point equidistance from its three sides.

Solution: In the adjoining figure, if O is the incentre of the triangle ABC, then OD = OE = OF [Inradius]

i.e. the point equidistance from three side of a triangle is incentre.

So the statement is false.

Example 3. In ΔABC, the internal bisectors ∠B and ∠C are meets at point O; if ∠BOC = 112°, then the value of ∠BAC is 44°.

Solution: In ΔABC, OB and OC are bisectors of ∠B and ∠C.

∴ ∠OBC = \(\frac{1}{2}\) ∠ABC and ∠OCB = \(\frac{1}{2}\) ∠ACB

∠OBC + ∠OCB = \(\frac{1}{2}\) (∠ABC + ∠ACB)

⇒ 180° – ∠BOC = \(\frac{1}{2}\) (180° – ∠BAC)

⇒ 180° – ∠BOC = 90° – \(\frac{1}{2}\) ∠BAC

⇒ 180° – 112° = 90° – \(\frac{1}{2}\) ∠BAC

⇒ 68° = 90° – \(\frac{1}{2}\) ∠BAC

⇒ \(\frac{1}{2}\) ∠BAC = 90° – 68° = 22°

⇒ BAC = 44°

∴ The statement is true.

Geometry Chapter 4 Theorems On Concurrence Fill In The Blanks

Example 1. The length of circumradius of a right-angled triangle is ________ of hypotenuse.

Solution: Half

[OA = OB = OC]

Example 2. The two medians of triangle are together than the third median.

Solution: Greater.

Example 3. If in ΔABC, three medians AD, BE and CF meets at point G then area of ΔABC area of ΔAGE is __________

Solution: ΔAGE = \(\frac{1}{6}\) ΔABC

⇒ \(\frac{\triangle \mathrm{ABC}}{\triangle \mathrm{AGE}}=\frac{6}{1}\)

⇒ ΔABC: ΔAGE = 6 : 1

Geometry Chapter 4 Theorems On Concurrence Short Answer Type Questions

Example 1. If the lengths of sides of triangle are 6 cm, 8 cm and 10 cm, then write where the circumcentre of this triangle lies.

Solution: 6²+ 8² = 100 = 10²

⇒ So the triangle is a right-angled triangle.

⇒ So the circumcentre of the triangle lies on the midpoint of hypotenuse i.e. lies on the midpoint of side with 10 cm in length.

Example 2. AD is the median and G is the centroid of an equilateral triangle. If the length of side 3√3 cm, then find the length of AG.

Solution: The medians and heights of any equilateral triangle are equal in length.

∴ The length of median AD.

= \(\frac{\sqrt{3}}{2}\) x length of side

= \(\left(\frac{\sqrt{3}}{2} \times 3 \sqrt{3}\right) \mathrm{cm}\)

= \(\frac{9}{2} \mathrm{~cm}=4.5 \mathrm{~cm}\)

centroid of a triangle divides any median from the vertex in the ratio 2: 1

∴ \(\frac{AB}{GD}\) = \(\frac{1}{2}\)

let AG = 2x cm

and GD x cm [x is common multiple and x > 0]

AG + GD = (2x + x) cm = 3x cm

3x = 4.5

⇒ x = 1.5

∴ AG = (2 x 1.5) cm = 3 cm.

Example 3. DEF is a pedal triangle of an equilateral triangle ABC. Find the value of ∠FDA.

Solution: In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°  [AD ⊥ BC]

⇒ hypotenuse AB = hypotenuse AC  [ΔABC is an equilateral]

⇒ and AD = AD [common side]

∴ ΔABD ≅ ΔACD [by R-H-S criterion of congruency]

∴ BD = CD i.e. D is the mid point of BC

and ∠BAD = ∠CAD = \(\frac{60^{\circ}}{2}\) = 30°

Similarly, F and E are the mid points of AB and AC respectively.

∴ FE || BC and FE = \(\frac{1}{2}\) BC

⇒ Similarly, DE = \(\frac{1}{2}\) AB and FD = \(\frac{1}{2}\) AC

As AB = BC = CA

∴ DE = FE = FD

∴ ΔDEF is an equilateral triangle.

∴ ∠ZDFE = 60°

⇒ As FE || BC and AB is intersection

∴ ∠AFE corresponding ∠ABC = 60°

∴ ∠AFD = ∠AFE + ∠DEF

= 60° + 60° = 120°

In ΔAFD, ∠FDA + ∠AFD + ∠FAD = 180°

∠FDA + 120° + 30 ° = 180°

⇒ ∠FDA = 30°

Example 4. ABC is an isosceles triangle in which ∠ABC = ∠ACB and median AD = \(\frac{1}{2}\) BC. If AB = √2 cm, then find the length of the circumradius of this triangle.

Solution: In ΔABC, ∠ABC = ∠ACB

∴ AC = AB

⇒ In ΔABD and ΔACD,

⇒ AB = AC, AD = AD  [common side]

⇒ and BD = CD  [D is the mid point of BC]

∴ ΔABD ≅ ΔACD [By S-S-S criterion of congruency]

∴ ∠ADB = ∠ADC

∠ADB + ∠ADC = 180°

∴ ∠ADB + ∠ADB = 180°

⇒ 2 ∠ADB = 180°

⇒ ∠ADB = 90°

∴∠ADC = 90°

⇒ Again, AD = \(\frac{1}{2}\) BC = BD = CD

∴ BD or CD or AD is the circumradius of ΔABC.

⇒ In right-angled triangle ABD, ∠ADB = 90°

∴ AD²+ BD² = AB² [By Pythagorus theorem]

BD² + BD² = (√2)² cm²

⇒ 2BD² = 2 cm²

⇒ BD² = 1 cm²

⇒ BD = √1 cm = 1 cm

⇒ The length of circumradius of ΔABC is 1 cm.

Example 5. In ΔABC, two medians AD and BE are perpendicular each other at point G. If BC = 8 cm and AC = 6 cm, then find the length of AB.

Solution: In ΔABC, the medians AD and BE intersects at G.

⇒ So G is the centroid of the ΔABC.

⇒ ∠AGB = ∠AGE = ∠DGB = 90° [AD ⊥ BE]

⇒ AG: GD = 2 : 1 and BG: GE = 2:1

⇒ Let, AG = 2x cm and GD = x cm

⇒ BG= 2y cm and GE = y cm

⇒ BG = 2y cm and GE = y cm [x and y are common multiples and x > 0, y > 0]

∴ BD = \(\frac{1}{2}\) BC = (\(\frac{1}{2}\) x 8) cm = 4 cm and AE = \(\frac{1}{2}\) AC = (\(\frac{1}{2}\) x 6) cm = 3 cm

⇒ In ΔAGE, ∠AGE = 90°

∴ AG² + GE² = AE² [By Pythagorus theorem]

⇒ (2x)² + (y)² = (3)²

⇒ 4x² + y² = 9………(1)

⇒ In ΔBGD, ∠BGD = 90°

∴ BG² + GD² = BD²

⇒ (2y)² + x² = 4²

⇒ 4y² + x² = 16 …….(2)

⇒ (1) + (2), we get,

⇒ 4x² + y² + 4y² + x² = 9 + 16

⇒ 5x² + 5y² = 25  ⇒ x² + y² = 5

⇒ In ΔABG, ∠AGB = 90°

∴ AB² = AG² + BG²

= ((2x)² + (2y)²) cm²

= 4(x² + y²) cm²

= 4 x 5 cm² = 20 cm²

⇒ AB = √20 cm

= √4×5 cm = 2√5 cm

⇒ The length of AB is 2√5 cm.

Example 6. O is the circumcentre of triangle ABC. If ∠OBC = 30° then find the value of ∠BAC.

Solution: I join A, O and AO is extended at P.

⇒ In ΔBOC, OB = OC [circumradius]

∴ ∠OCB = ∠OBC = 30°

⇒ ∠BOC = 180° (30° + 30°) = 120°

⇒ In ΔAOB, OA = OB

∴ ∠OAB = ∠OAB

The exterior ∠BOP = ∠OAB + ∠OBA

= ∠OAB + ∠OAB = 2 ∠OAB

Similarly, ∠COP = 2 ∠OAC

∠BAC = ∠OAB + ∠OAC

= \(\frac{1}{2}\)(∠BOP + ∠COP)

= \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) x 120° = 60°

Example 7. If the point O is the orthocentre of ΔABC then find the relation between ∠BOC and ∠BAC.

Solution: In ΔABC,

⇒ AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.

⇒ AD, BE and CF intersects at point O.

∴ ∠AEB = ∠AFC = 90°

⇒ In quadrilateral AEOF,

⇒ ∠EAF + ∠AFO + ∠EOF + ∠AEO = 360°

⇒ ∠EAF + 90° + ∠EOF + 90° = 360°

⇒ ∠EAF + ∠BOC = 360° – 180°  [∠EOF = vertically opposite ∠BOC]

⇒ i.e. ∠BAC + ∠BOC = 180° [required relation]

Example 8. In ΔABC, D, E and F are midpoints of side, BC, CA and AB respectively. If AB = 5 cm, BC = 6 cm and CA = 7 cm, then find the perimeter of ΔDEF.

Solution: In ΔABC, F and E are midpoints of side AB and AC respectively,

∴ FE = \(\frac{1}{2}\) BC = (\(\frac{1}{2}\) x 6) cm = 3 cm

⇒ Similarly, DE = \(\frac{1}{2}\) = AB = (\(\frac{1}{2}\) x 5) cm = 2.5 cm

⇒ and FD = \(\frac{1}{2}\) AC = (\(\frac{1}{2}\) x 7) cm = 3.5 cm

∴ Perimeter of ΔDEF = (3 + 2.5+ 3.5) cm = 9 cm.