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(1) Prove that sum of the angles of a quadrilateral is .
Given: Quadrilateral ABCD
To Prove:
Construction: Join ACProof: In , We have
..........(i)
In , we have
...............(ii)
Adding (i) and (ii), we get
(2) Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Given: A parallelogram ABCD
To Prove: A diagonal, say, AC, of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e.
Construction: Join ACProof: Since ABCD is a parallelogram. Therefore,
and
Now, and transversal AC intersects them at A and C respectively.
â€¦â€¦.(i)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Alternate interior angles]
Again, and transversal AC intersects them at A and C respectively. Therefore,
Â â€¦â€¦(ii) Â Â Â Â Â Â Â Â [Alternate interior angles]
Now, in ABC and CDA, we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (i)]
So, by ASA congruence criterion, we have
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(3) Prove that two opposite angles of a parallelogram are equal.
Given: A parallelogram ABCD
To prove: and Proof: Since ABCD is a parallelogram. Therefore,
and
Now, and transversal AD intersects them at A and D respectively.
Â â€¦â€¦.(i)Â Â [Sum of Consecutive interior anglesis ]
Again, and DC intersects them at D and C respectively.
â€¦.. (ii)Â Â Â Â [Sum of Consecutive interior angles is ]
From (i) and (ii), we get
.
Similarly, .
Hence, and
(4) Prove that the diagonals of a parallelogram bisect each other.
Given: A parallelogram ABCD such that its diagonals AC and BD intersect at O.
To prove: and Proof: Since ABCD is a parallelogram. Therefore,
and
Now, and transversal AC intersects them at A and C respectively.
â€¦â€¦..(i)
Again, and BD intersects them at B and D respectively.
â€¦â€¦..(ii)
Now, in AOB and COD, we have
and,
So, by ASA congruence criterion
and
Hence, and
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(5) Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
Given: A parallelogram ABCD such that the bisectors of consecutive angles A and B intersect at P.
To prove: Proof: Since ABCD is a parallelogram. Therefore,
Now, and transversal AB intersects them.
â€¦.(i)
AP is the bisector of and BP is the bisector of then Â and
In , we have
Â Â Â Â Â [From (i)]
(6) Prove that if a diagonal of a parallelogram bisects one of the angles of the parallelogram it also bisects the second angle.
Given: A parallelogram ABCD in which diagonal AC bisects .
To prove: AC bisects Proof: Since ABCD is a parallelogram. Therefore,
Now, and AC intersects them.
Â Â â€¦â€¦(i) [Alternate interior angles]
Again, and AC intersects them.
Â â€¦â€¦(ii) [Alternate interior angles]
But, it is given that AC is the bisector of . Therefore,
Â Â Â Â Â Â Â Â â€¦â€¦..(iii)
From (i), (ii) and (iii), we get
Â Â Â Â Â Â Â â€¦â€¦â€¦(iv)
Hence, AC bisects .
From (ii) and (iii), we have
Â Â Â Â Â [Angles opposite to equal sides are equal]
But, and Â Â Â [ ABCD is a parallelogram]
Hence, ABCD is a rhombus.
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(7) Prove that the angles bisectors of a parallelogram form a rectangle.
Proof: Since ABCD is a parallelogram. Therefore,
Now, and transversal AB intersects them at A and B respectively. Therefore,
Â [Sum of consecutive interior angles is ]
Â â€¦.(i)Â Â [AS and BS are bisectors of and respectively]
But, in , we have
[Sum of the angle of a triangle is ]
[ and are vertically opposite angles ]
Similarly, we can prove that
, and
Hence, PQRS is a rectangle.
(8) Prove that a quadrilateral is a parallelogram if its opposite sides are equal.
Given: A quadrilateral ABCD in which and
To prove: ABCD is a parallelogram.
Construction: Join AC.Proof: In ACB and CAD, we have
Â Â [Common Side]
So, by SAS criterion of congruence, we have
ACB and CAD
Â Â Â Â â€¦.(i)
And,
Now, line AC intersects AB and DC at A and C, such that
Â Â Â .....(ii)
i.e., alternate interior angles are equal.
Â Â Â â€¦..(iii)
Similarly, line AC intersects BC and AD at C and A such that
i.e., alternate interior angles are equal.
Â Â Â â€¦..(iv)
From (iii) and (iv), we have
and
Hence, ABCD is a parallelogram.
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(9) Prove that a quadrilateral is a parallelogram if its opposite angles are equal.
Given: A quadrilateral ABCD in which and .
To prove: ABCD is a parallelogram.Proof: In quadrilateral ABCD, we have
â€¦â€¦(i)
Â Â â€¦â€¦(ii)
Â â€¦..(iii)
Since sum of the angles of a quadrilateral is
Â â€¦â€¦(iv)
Â â€¦..(v) Â Â []
Now, line AB intersects AD and BC at A and B respectively such that
i.e. the sum of consecutive interior angles is
Â Â Â â€¦â€¦(vi)
Again,
Now, line BC intersects AB and DC at A and C respectively such that
i.e., the sum of consecutive interior angles is .
Â Â Â Â â€¦â€¦(vii)
From (vi) and (vii), we get
and .
Hence, ABCD is a parallelogram.
(10) Prove that the diagonals of a quadrilateral bisect each other, if it is a parallelogram.
Proof: In AOD and COB, we have
So, by SAS criterion of congruence, we have
Now, line AC intersects AD and BC at A and C respectively such that
i.e., alternate interior angles are equal.
Similarly,
Hence, ABCD is a parallelogram.
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(11) Prove that a quadrilateral is a parallelogram if its one Pair of opposite sides are equal and parallel.
Given: A quadrilateral ABCD in which and .
To prove: is a parallelogram.
Construction: Join AC.Proof: In ABC and CDA, we have
And,
So, by SAS criterion of congruence, we have
Thus, line AC intersects AB and DC at A and C respectively such that
i.e., alternate interior angles are equal.
.
Thus, and .
Hence, quadrilateral ABCD is a parallelogram.
(12) Prove that each of the four angles of a rectangle is a right angle.
Given: A rectangle ABCD such that
To prove: Proof: Since ABCD is a rectangle.
ABCD is a parallelogram
Now, and line AB intersects them at A and B.
Similarly, we can show that and
Hence,
(13) Prove that each of the four sides of a rhombus of the same length.
Given: A rhombus ABCD such that .
To prove: .Proof: Since ABCD is rhombus
ABCD is a parallelogram
and
But,
Hence, all the four sides of a rhombus are equal.
(14) Prove that the diagonals of a rectangle are of equal length.
Given: A rectangle ABCD with AC and BD as its diagonals.
To prove: .Proof: Since ABCD is a rectangle
ABCD is a parallelogram such that one of its angles, say , is a right angle.
and
Now, and AB intersects them at A and B respectively.
]
In ABD and BAC, we have
And,
So, by SAS criterion of congruence, we have
}
Hence,
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(15) Prove that diagonals of a parallelogram are equal if and only if it is a rectangle.
Proof: In ABC and DCB, we have
And, So, by SAS criterion of congruence, we have
But, and BC cuts them.
Thus, .
Hence, ABCD is a rectangle.
(16) Prove that the diagonals of a rhombus are perpendicular to each other.
Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
To prove: 4Proof: We know that a parallelogram is a rhombus, if ll of its sides are equal. So, ABCD is a rhombus. This implies that ABCD is a parallelogram such that
Â Â â€¦..(i)
Since the diagonals of a parallelogram bisect each other.
and Â Â Â â€¦..(ii)
Now, in BOC and DOC, we have
So, by SSS criterion of congruence, we have
But,
Similarly,
Hence,
(17) Prove that diagonals of a parallelogram are perpendicular if and only if it is a rhombus.
Given: A parallelogram ABCD in which .
To prove: Parallelogram ABCD is a rhombus.Proof: Suppose AC and BD intersect at O. Since the diagonals of a parallelogram bisect each other. So, we have
Â Â â€¦â€¦(i)
Now, in AOD and COD, we have
So, by SAS criterion of congruence, we have
Â Â Â â€¦â€¦(ii)
Since ABCD is a parallelogram.
and
Hence, parallelogram ABCD is a rhombus.
(18) Prove that the diagonals of a square are equal and perpendicular to each other.
Given: A square ABCD.
To prove: and .Proof: In ADB and BCA, we have
And,
So, by SAS criterion of congruence, we have
Now, in AOB and AOD, we have
And,
So, by SSS criterion of congruence, we have
But,
Hence, and
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(19) Prove that if the diagonals of a parallelogram are equal and intersect at right angles, Â then it is square
Given: A parallelogram ABCD in which and
To prove: ABCD is a square.Proof: In AOB and AOD, we have
And
So, by SAS criterion of congruence, we have
But, and
Â Â â€¦â€¦.(i)
Now, in ABD and BAC, we have
And,
So, by SSS criterion of congruence, we have
But,
Â Â â€¦â€¦(ii)
From (i) and (ii), we obtain that ABCD is a parallelogram whose all side are equal and all angles are right angles.
Hence, ABCD is a square.
(20) Prove that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
Given: A in which D and E are the mid points of sides AB and AC respectively. DE is joined
To prove: and
Construction: Produce the line segment DE to F, such that . Join FCProof: In AED and CEF, we have
And,
So, by SAS criterion of congruence, we have
Â Â Â â€¦(i)
And, Â â€¦â€¦(ii)
Now, D is the mid-point of AB
Â Â â€¦..(iii)
Now, DF intersects AD and FC at D and F respectively such that
i.e. alternate interior angles are equal.
Â Â Â â€¦..(iv)
From (iii) and (iv), we find that DBCF is a quadrilateral such that one pair of sides are equal and parallel.
DBCF is a parallelogram.
and
But, D,E,F are collinear and.
and
(21) Prove that a line through the mid-point of a side of a triangle parallel to another side bisects theÂ third side.
Proof: We have to prove that E is the mid-point of AC. If possible, let E be not the mid-point of AC. Let E prime be the mid-point AC. Join DE prime.Now, in , D is the mid-point of AB and E prime is the mid-point of AC. We have,
Â Â â€¦..(i)
Also, Â â€¦.(ii)
From (i) and (ii), we find that two intersecting lines DE and DEâ€™ are both parallel to Line BC.
This is contradiction to the parallel line axiom.
So, our supposition is wrong. Hence, E is the mid-point of AC.
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(22) Prove that the quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.
Given: ABCD is a quadrilateral in which P,Q,R,S are the mid-points of sides AB, BC, CD and DA respectively.
To prove: PQRS is a parallelogram.Proof: In , P and Q are the mid-points of sides AB and BC respectively.
and Â Â â€¦.(i)
In , R and S are the mid-points of CD and AD respectively.
and Â Â Â â€¦â€¦.(ii)
From (i) and (ii), we have
and
Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.
Hence, PQRS is a parallelogram.
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