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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.
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

1 comment:

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