Tuesday, 5 July 2011

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: mQPX + mYPQ = 180°.
Construction: Draw PE perpendicular to XY.
Proof: mQPX = mQPE + mEPX
= mQPE + 90° (1)
mYPQ = mYPE − mQPE
= 90° − mQPE (2)
(1) + (2) mQPX + mYPQ = (mQPE + 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: mAOC = mBOD, mBOC = mAOD.
Proof: The ray OB stands on the line CD.

mBOD + mBOC = 180° (1)
The ray OC stands on the line AB.
mBOC + mAOC = 180° (2)
From (1) and (2),
mBOD + mBOC = mBOC + mAOC
mBOD = mAOC.
Since the ray OA stands on the line CD,
mAOC + mAOD = 180° (3)
From (2) and (3), we get
mBOC + mAOC = mAOC + mAOD
mBOC = mAOD.
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 ∴ mXAB = mABC (alternate angles).
= B. (1)
Next, AC is a transversal to the parallel lines XY and BC.
mYAC = mACB (alternate angles) = C. (2)
We also have mBAC = mA. (3)
(1) + (2) + (3) mXAB + mYAC + mBAC = mB + mC + mA
(mXAB + mBAC) + mCAY = mA + mB + mC
mXAC + mCAY = mA + mB + mC
180° = mA + mB + mC.
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 mB > mC.

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, mC > mB. This contradicts the given fact that mB > mC. 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, mBAD = mDAC 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.
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.
(angles opposite to the equal sides are equal) (2)
From (1) and (2), we get
i.e., AC bisects the angle A.
Similarly we can prove that AC bisects C, BD bisects B and BD bisects D


  1. Your blog contains all the article which are really important in students studies & also it covers all the detailed points & also they are very easy to understand & very helpful

  2. i asked for circle not for triangles and quadrilaterals theorem.

  3. For chapter circle not for triangles...

  4. Really useless every one knows all these

  5. I asked for area of parallelogram and triangles not of quadrilateral

    1. Area og parallelogram = base × height
      Area of triangle = base×height(if triangle is equilateral or isoceles)
      Area of right anle triangle= 1÷2×base×height(if trianle is right angle)

  6. These theorems are not of ant use in circle chaptet.This does not contain NCERT book theorem.You made ur own theorems for circle chapter.Unsatisfie,very dissatisfying.Disappoined!:( :(