 Theorem 1 -Angle sum property of Quadrilateral:

The sum of four angles of a Quadrilateral is $\text{360°}$ Given: A Quadrilateral $\text{ABCD}$.

To prove :

Construction: Join AC

Proof: In $\Delta \text{ABC}$

(By angle sum property of triangle)

In $\Delta \text{ADC}$

(By angle sum property of triangle)

Remarks:

·            A parallelogram is a trapezium. But a trapezium is not parallelogram.

·            A rectangle is parallelogram, But a parallelogram is not a rectangle.

·            A rhombus is parallelogram, but parallelogram is not a rhombus.

·            A square is parallelogram, rectangle, rhombus but not viceversa.

·            A kite is not a parallelogram.

Properties of parallelogram:

A Quadrilateral is parallelogram if both the pairs of opposite sides are parallel.

Theorem 2:

A diagonal of a parallelogram divides it into two congruent triangles.

Given:

$\text{ABCD}$ is parallelogram. $\text{AC}$ is diagonal To prove:

$\Delta \text{ABC}\cong \Delta \text{ADC}$

Proof:

$\text{AD}\parallel \text{BC}$

And $\text{AC}$ is transversal

So …..(Alternate angles)

$\text{AB}\parallel \text{DC}$ and $\text{AC}$ is transversal

……(Alternate angles)

…(Common)

$\Delta \text{ABC}\cong \Delta \text{CDA}$ …( By ASA rule.)

Theorem 3:

In a parallelogram, opposite sides are equal and opposite angles are equal.

Given :

$\text{ABCD}$ is parallelogram To prove:

and

Construction: Join AC

Proof:

$\text{AD}\parallel \text{BC}$

And $\text{AC}$ is transversal

…..(Alternate angles)

$\text{AB}\parallel \text{DC}$ and AC is transversal

……(Alternate angles)

…(Common)

$\Delta \text{ABC}\cong \Delta \text{CDA}$ (By ASA rule.)

(CPCT).

Similarly it can be proved that

Theorem 4:

The diagonals of a parallelogrambisect each other.

Given: $\text{ABCD}$ is a parallelogram

$\text{AB}\parallel \text{CD}$ Ac and $\text{BD}$ intersect at $\text{O}$.

To prove:

$\text{O}$ is midpoint of $\text{AC}$ and BD.

That is

Proof:

In $\Delta \text{AOB}$ and $\Delta \text{DOC}$

….( Alternate angles, $\text{AB}\parallel \text{CD}$,BD as transversal)

…. ( Alternate angles, $\text{AB}\parallel \text{CD}$,AC as transversal)

…… (Opp sides of parallelogram)

$\Delta \text{OBA}\cong \Delta \text{ODC}$ …..(By ASA rule.)

So (CPCT)

Theorem 5:

A Quadrilateral is parallelograms if opposite sides are congruent.

Given:

$\text{ABCD}$ is Quadrilateral With and

To prove: ABCD is parallelogram

i.e.

$\text{AB}\parallel \text{CD}$

$\text{AD}\parallel \text{BC}$

Construction:Join $\text{BD}$

Proof:

In $\Delta \text{ABD}$ and $\Delta \text{CDB}$

….(Given)

….(Given)

…(Common)

$\Delta \text{ABD}\cong \Delta \text{CDB}$ ….(SSS rule)

….. (CPCT)

$\text{AB}\parallel \text{CD}$ …..(Alternate angle test)

Similarly

(CPCT)

$\text{AD}\parallel \text{BC}$ …(Alternate angle test)

AS both the pair of opposite sides areparallel, we say $\text{ABCD}$ is parallelogram.

Theorem 6:

A Quadrilateral is parallelogram if opposite angles are congruent.

Given:

$\text{PQRS}$ is Quadrilateral

With and To prove: $\text{PQRS}$ is parallelogram

i.e.

$\text{PQ}\parallel \text{RS}$

$\text{PS}\parallel \text{QR}$

Construction : Join $\text{PR}$

Proof:

So …(1)

But we know

(From $\text{1}$ and $\text{2}$ )

$\therefore \text{PQ}\parallel \text{RS}$ ( Sum of interior angle criteria)

Similarly,

$\text{PS}\parallel \text{QR}$ can be proved.

As both the pair of sides are parallel, we say $\text{PQRS}$ is parallelogram.

Theorem 7:

A Quadrilateral is parallelogram if a pair of opposite sides is congruent and parallel.

Given: $\text{ABCD}$ is quadrilateral.

$\text{AB}\parallel \text{CD}$

To prove: ABCD is parallelogram.

i.e. $\text{BC}\parallel \text{AD}$

Proof: Construction: join AC

Proof:

In $\Delta \text{ADC}$ and $\Delta \text{CBA}$

…(Given)

…(Alternate angle as $\text{AB}\parallel \text{CD}$ )

…..(Common)

$\Delta \text{ADC}\cong \Delta \text{CBA}$ … (SAS rule)

So ….( CPCT)

$\text{BC}\parallel \text{AD}$ (Alternate angle criteria)

Converse :

Theorem 8 : If diagonals of parallelogram are equal then it is rectangle.

Given:

$\text{XYZW}$ is parallelogram

be its diagonal. To prove: $\text{XYZW}$ is rectangle

i.e.

Proof:

In $\Delta \text{XWZ}$ and $\Delta \text{YZW}$

….(Given)

….(Opp. side of parallelogram)

…(Common)

$\Delta \text{XWZ}\cong \Delta \text{YZW}$ …(SSS rule)

(CPCT)…..1

But …(Interior angles are supplementary)

…(From 1)

Similarly we can show all the angles of quadrilateral are $\text{90°}$

Hence it is rectangle.

Theorem 9:

Diagonals of a rhombus are perpendicular to each other.

Given :

$\text{ABCD}$ is rhombus.

$\therefore \text{ABCD}$ is parallelogram. To prove:

$\text{AC}\perp \text{BD}$

Proof:

In $\Delta \text{AOB}$ and $\Delta \text{BOC}$

… .(Given)

……. (Diagonals of parallelogram bisect each other)

……..( Common)

$\Delta \text{AOB}\cong \Delta \text{COB}$ …(SSS test)

So …(CPCT)

But ….(Linear pair)

$\therefore \text{AC}\perp \text{BD}$

Theorem 10- Mid-point theorem:

The line segment joining the midpoint of any two sides of a triangle is parallel to third side and is half of third side.

Given: In $\Delta \text{ABC}$ $\text{D}$ and $\text{E}$ are midpoints of the sides $\text{AB}$ and $\text{AC}$ respectively. To prove:

1) $\text{DE}\parallel \text{BC}$ 2) .

Construction: produce DE so that DE=EF. Join CF Proof:

In $\Delta \text{ADE}$ and $\Delta \text{CEF}$

…. (E is midpoint of AC)

….. (By construction)

……(V. opp angles)

$\Delta \text{ADE}\cong \Delta \text{CEF}$ …… (SAS rule)

(CPCT)

……. (CPCT)

But …….(D is midpoint)

So

(CPCT)

$\text{AD}\parallel \text{FC}$

$\square \text{DBCF}$ is parallelogram…… (One pair of side is parallel and equal)

So, $\text{DF}\parallel \text{BC}$

(By construction)

Theorem 11 -Converse of midpoint theorem:

The line through the midpoint of one side of a triangle, parallel to other side, intersects the third side at its midpoint.

Given:

In $\Delta \text{ABC}$, $\text{D}$ is mid-point of $\text{AB}$

$\text{DE}\parallel \text{BC}$ To prove:

$\text{E}$ is midpoint of $\text{AC}$.

Proof:

Suppose $\text{E}$ is not midpoint of $\text{AC}$.

Let $\text{E}$ ’ is midpoint of $\text{AC}$. Then By mid-point theorem, $\text{DE'}\parallel \text{BC}$

But $\text{DE}\parallel \text{BC}$ …….. (Given)

$\text{DE}$ and $\text{DE}$ ’ are two intersecting line are parallel to $\text{BC}$.

So $\text{E}$ is mid-point of $\text{AC}$.