CBSE NCERT Maths IX Theorem Proved

 Theorems with logical proofs

Theorem 1. : If a ray stands on a line, then the sum of the adjacent angles so formed is 180°.

Given: The ray PQ stands on the line XY.

To Prove: m∠QPX + m∠YPQ = 180°.

Construction: Draw PE perpendicular to XY.

Proof: m∠QPX = m∠QPE + m∠EPX

= m∠QPE + 90° (1)

m∠YPQ = m∠YPE − m∠QPE

= 90° − m∠QPE (2)

(1) + (2) ⇒ m∠QPX + m∠YPQ = (m∠QPE + 90°) + (90° − m ∠QPE) = 180°.

Thus the theorem is proved.

Theorem 2: If two lines intersect, then the vertically opposite angles are of equal measure.

Given: Two lines AB and CD intersect at the point O

To prove: m∠AOC = m∠BOD, m∠BOC = m∠AOD.

Proof: The ray OB stands on the line CD.

∴ m∠BOD + m∠BOC = 180° (1)

The ray OC stands on the line AB.

∴ m∠BOC + m∠AOC = 180° (2)

From (1) and (2),

m∠BOD + m∠BOC = m∠BOC + m∠AOC

∴ m∠BOD = m∠AOC.

Since the ray OA stands on the line CD,

m∠AOC + m∠AOD = 180° (3)

From (2) and (3), we get

m∠BOC + m∠AOC = m∠AOC + m∠AOD

∴ m∠BOC = m∠AOD.

Hence the theorem is proved.

Theorem 3: The sum of the three angles of a triangle is 180°.

Given: ABC is a triangle (see Figure 6.54).

To prove: ∠A + ∠B + ∠C = 180°.

Construction: Through the vertex A, draw the line

XY parallel to the side BC.

Proof: XY || BC ∴ m∠XAB = m∠ABC (alternate angles).

= ∠B. (1)

Next, AC is a transversal to the parallel lines XY and BC.

∴ m∠YAC = m∠ACB (alternate angles) = ∠C. (2)

We also have m∠BAC = m∠A. (3)

(1) + (2) + (3) ⇒ m∠XAB + m∠YAC + m∠BAC = m∠B + m∠C + m∠A

⇒ (m∠XAB + m∠BAC) + m∠CAY = m∠A + m∠B + m∠C

⇒ m∠XAC + m∠CAY = m∠A + m∠B + m∠C

⇒ 180° = m∠A + m∠B + m∠C.

Hence the theorem is proved.

Theorem4. : The angles opposite to equal sides of a triangle are equal.

Given: ABC is a triangle where AB = AC).

To prove: ∠B = ∠C.

Construction: Mark the mid point of BC as M and join AM.

Proof: In the triangles AMB and AMC

(i) BM = CM (ii) AB = AC (iii) AM is common.

∴ By the SSS criterion, ΔAMB ≡ ΔAMC.

∴ Corresponding angles are equal. In particular, ∠B = ∠C.

Hence the theorem is proved.

Theorem 5. : The side opposite to the larger of two angles in a triangle is longer than the side opposite to the smaller angle.

Given: ABC is a triangle, where ∠B is larger than ∠C, that is m∠B > m∠C.

To prove: The length of the side AC is longer than the length of the side AB.

i.e., AC > AB (see Figure 6.56).

Proof: The lengths of AB and AC are positive numbers. So three cases arise

(i) AC < AB

(ii) AC = AB

(iii) AC > AB

Case (i) Suppose that AC < AB. Then the side AB has longer length than the side AC. So the angle ∠C which is opposite to AB is larger measure than that of ∠B which is opposite to the shorter side AC. That is, m∠C > m∠B. This contradicts the given fact that m∠B > m∠C. Hence the assumption that AC < AB is wrong.

∴AC < AB.

Case (ii) Suppose that AC = AB. Then the two sides AB and AC are equal. So the angles opposite to these sides are equal. That is ∠B = ∠C. This is again a contradiction to the given fact that ∠B > ∠C. Hence AC = AB is impossible. Now Case (iii) remains alone to be true.

Hence the theorem is proved.

Theorem 6. : A parallelogram is a rhombus if its diagonals are perpendicular.

Given: ABCD is a parallelogram where the diagonals AC and BD are perpendicular.

To prove: ABCD is a rhombus.

Construction: Draw the diagonals AC and BD. Let M be the point of intersection of AC and BD (see Proof: In triangles AMB and BMC,

(i) ∠AMB = ∠BMC = 90°

(ii) AM = MC

(iii) BM is common.

∴ By SSA criterion, ΔAMB ≡ ΔBMC.

∴Corresponding sides are equal.

In particular, AB = BC.

Since ABCD is a parallelogram, AB = CD, BC = AD.

∴ AB = BC = CD = AD.

Hence ABCD is a rhombus. The theorem is proved.

Theorem 7:Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base.

Solution: Let ABC be an isosceles triangle where AB = AC. Let AD be the bisector of the vertex angle ∠A. We have to prove that AD is the median of the base BC. That is, we have to prove that D is the mid point of BC. In the triangles ADB and ADC,

we have AB = AC, m∠BAD = m∠DAC AD is an angle (bisector), AD is common.

∴ By SAS criterion, ∠ABD ≡ ΔACD.

∴ The corresponding sides are equal.

∴ BD = DC.

i.e., D is the mid point of BC.

Theorem 8 Prove that the sum of the four angles of a quadrilateral is 360°.

Solution: Let ABCD be the given quadrilateral. We have to prove that ∠A + ∠B + ∠C + ∠D = 360°. Draw the diagonal AC. From the triangles ACD and ABC, we get

∠DAC + ∠D + ∠ACD = 180° (1)

∠CAB + ∠B + ∠ACB = 180° (2)

(1) + (2) ⇒ ∠DAC + ∠D + ∠ACD + ∠CAB

+ ∠B + ∠ACB = 360°

⇒ (∠DAC + ∠CAB) + ∠B + (∠ACD + ∠ACB)+ ∠D = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°.

Theorem 8: In a rhombus, prove that the diagonals bisect each other at right angles.

Solution: Let ABCD be a rhombus, Draw the diagonals AC and BD. Let them meet at O. We have to prove that O is the mid point of both AC and BD and that AC is perpendicular (⊥) to BD.

Since a rhombus is a parallelogram, the diagonals AC and BD bisect each other.

∴ OA = OC, OB = OD.

In triangles AOB and BOC, we have

(i) AB = BC

(ii) OB is common

(iii) OA = OC

∴ ΔAOB = ΔBOC, by SSS criterion.

∴ ∠AOB = ∠BOC.

Similarly, we can get ∠BOC = ∠COD, ∠COD = ∠DOA.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = x (say)

But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

∴ x + x + x +x = 360°

∴ 4x = 360° or x = 4360° = 90°.

∴ The diagonals bisect each other at right angles.

Theorem 9: Prove that a diagonal of a rhombus bisects each vertex angles through which it passes.

Solution: Let ABCD be the given rhombus. Draw the diagonals AC and BD. Since AB || CD and AC is a transversal to AB and CD. We get

∠BAC = ∠ACD (alternate angles are equal) (1)

But AD = CD (since ABCD is a rhombus)

∴ΔADC is isosceles.

∴∠ACD = ∠DAC

(angles opposite to the equal sides are equal) (2)

From (1) and (2), we get

∠BAC = ∠DAC

i.e., AC bisects the angle ∠A.

Similarly we can prove that AC bisects ∠C, BD bisects ∠B and BD bisects ∠D.