Characteristics Of Parallelogram

Theorem: A quadrilateral is a parallelogram if and only if any one of the following statements is true:
(i) Each diagonal divides the quadrilateral into two congruent triangles.
(ii) Two pairs of opposite sides are equal in length.
(iii) Two pairs of opposite angles are equal in measure.
(iv) Adjacent angles are supplementary.
(v) The diagonals bisect each other.
(vi) One pair of opposite sides are parallel and equal in length.

Prerequisites:
Parallelogram (definition)
Vertical angle theorem (proof)
Alternate angles property (proof)
Interior angle property of parallel lines (proof)
ASA congruency (proof)
SSS congruency (proof)
SAS congruency (proof)
Angle sum property of a polygon (proof)

Proof:

(i) Let ABCD be a parallelogram with AC being one of its diagonal.

In the triangles $\triangle ABC$ and $\triangle ADC$,

$\qquad\quad\angle BAC = \angle ACD\qquad\qquad\qquad\qquad\text{(Alternate angles as $AB\parallel DC$)}\\
\qquad\quad\angle BCA = \angle CAD\qquad\qquad\qquad\qquad\text{(Alternate angles as $AD\parallel BC$)}\\
\text{Also,}\quad AC = AC\qquad\qquad\qquad\qquad\qquad\;\;\;\text{(common in both triangles)}$

Hence, $\triangle ABC\cong\triangle ADC$ by ASA congruency rule.

Conversely, if $\triangle ABC\cong\triangle ADC$, then by CPCTC,

$\qquad\quad\angle CAB = \angle ACD\quad\Rightarrow AB \parallel CD\qquad\qquad\qquad\text{(by alternate angle property)}\\
\qquad\quad\angle ACB = \angle CAD\quad\Rightarrow AD \parallel BC\qquad\qquad\qquad\text{(by alternate angle property)}$

Hence, ABCD is a parallelogram (see definition of a parallelogram).

(ii) Let ABCD be a parallelogram. Since $\triangle ABC\cong\triangle ADC$, therefore by CPCTC, side AB $=$ CD $\quad\;\;\:\!$and BC $=$ AD.

Conversely, if AB $=$ CD and BC $=$ AD, then in $\triangle ABC$ and $\triangle ADC$,

$\qquad\quad AB = CD\qquad\qquad\qquad\qquad\text{(by assumption)}\\
\qquad\quad BC = AD\qquad\qquad\qquad\qquad\text{(by assumption)}\\
\qquad\quad AC = AC\qquad\qquad\qquad\qquad\text{(common)}$

Hence, $\triangle ABC\cong\triangle ADC$ by SSS congruency rule.
Therefore by (i), ABCD is a parallelogram

(iii) Let ABCD be a parallelogram. Since $\triangle ABC\cong\triangle ADC$, therefore by CPCTC, $\angle B = \angle D$ $\qquad$and $\angle A = \angle C$.

Conversely, let $\angle B = \angle D$, and $\angle A = \angle C$.

$\qquad\quad\angle A + \angle B + \angle C + \angle D = 360^o\qquad\!\qquad\text{(by angle sum property of a polygon)}\\
\therefore\quad\;\;\; 2\angle A + 2\angle B = 360^o\qquad\qquad\qquad\qquad\;\;\text{(as $\angle C = \angle A$ and $\angle D = \angle B$)}\\
\Rightarrow\quad\;\; \angle A + \angle B = 180^o\\
\Rightarrow\quad\;\; AD \parallel BC\qquad\qquad\qquad\qquad\qquad\;\;\;\qquad\text{(by interior angle property of parallel lines)}$

Similarly, $AB \parallel CD$.

Hence, ABCD is a parallelogram.

(iv) If ABCD is a parallelogram, then since AB $\parallel$ CD, hence by interior angle property of parallel $\qquad\!$lines, angles $\angle A$ and $\angle D$ are supplementary. Similarly, all other pairs of adjescent angles $\qquad\!$are supplementary.

$\qquad\!$Conversely, if $\angle A$ and $\angle D$ are supplementary, then by interior angle property, AB $\parallel$ CD. $\qquad\!$Similarly, AD $\parallel$ BC. Hence, ABCD is a parallelogram.

(v) Let ABCD be a parallelogram. Let the diagonals AC and BD intersect at E.

In $\triangle ADE$ and $\triangle CBE$,

$\qquad\quad\angle DAE = \angle BCE\qquad\qquad\qquad\qquad\text{(Alternate angles as $AD\parallel BC$)}\\
\qquad\quad\angle ADE = \angle CBE\qquad\qquad\qquad\qquad\text{(Alternate angles as $AD\parallel BC$)}\\
\text{Also,}\quad AD = BC\qquad\qquad\qquad\qquad\qquad\;\;\:\!\text{(from (ii))}$

Hence, $\triangle ADE\cong\triangle CBE$ by ASA congruency rule. Thus, by CPCTC:

$\qquad\quad AE = CE\\
\qquad\quad DE = BE$

Hence, E is the midpoint of both AC and BD, i.e., both the diagonals bisect each other.

Conversely, let diagonals bisect each other, then in $\triangle ADE$ and $\triangle CBE$,

$\qquad\quad AE = CE \qquad\qquad\qquad\qquad\text{(as E is the midpoint of AC)}\\
\qquad\quad DE = BE \qquad\qquad\qquad\qquad\text{(as E is the midpoint of BD)}\\
\qquad\quad\angle AED = \angle CEB\qquad\qquad\quad\text{(Vertically opposite angles)}$

Hence, $\triangle ADE\cong\triangle CBE$ by SAS congruency rule.

Therefore, by CPCTC:

$\qquad\quad\angle EAD = \angle ECB\quad\Rightarrow AD \parallel BC\qquad\qquad\text{(by alternate angle property)}$

Similarly, by proving $\triangle ABE\cong\triangle CDE$, line AB can be shown to be parallel to line CD.

Hence, ABCD is a parallelogram.

(vi) If ABCD is a parallelogram, then AD $\parallel$ BC (by definition of parallelogram) and AD $=$ BC (by $\qquad\!$(ii)).

Conversely, if AD $\parallel$ BC and AD $=$ BC, then in $\triangle ADE$ and $\triangle CBE$,

$\qquad\quad AD = BC\qquad\qquad\qquad\qquad\qquad\;\;\;\text{(given)}\\
\qquad\quad\angle DAE = \angle BCE\qquad\qquad\qquad\qquad\text{(alternate angles)}\\
\qquad\quad\angle ADE = \angle CBE\qquad\qquad\qquad\qquad\text{(alternate angles)}$

Hence, $\triangle ADE\cong\triangle CBE$ by ASA congruency rule. Thus, by CPCTC:

$\qquad\quad AE = CE\\
\qquad\quad DE = BE$

Hence, both the diagonals bisect each other. Thus, by (v), ABCD is a parallelogram.

Q.E.D.


Recommended:
Diagonal property of rectangle
Properties of rhombus
Angle sum property of polygon