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**(1) Prove that sum of the angles of a quadrilateral is .
**

..........(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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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.
**

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
**

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.
**

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.
**

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.
**

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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