NCERT Ganita Manjari Part II • Parallelograms, Mid-Point & Varignon's Theorem
1. Introduction & Angle Sum Property of a Quadrilateral
A closed four-sided polygon lying in a flat plane is called a quadrilateral. It consists of 4 vertices, 4 sides, and 4 interior angles.
$$\text{Angle Sum Property: } \angle A + \angle B + \angle C + \angle D = 360^\circ$$
Proof: Draw diagonal $AC$ dividing quadrilateral $ABCD$ into two triangles $\triangle ABC$ and $\triangle ADC$. The sum of angles in $\triangle ABC = 180^\circ$ and in $\triangle ADC = 180^\circ$. Adding both sums yields $180^\circ + 180^\circ = 360^\circ$.
2. Classification & Family Tree of Quadrilaterals
Type
Defining Feature
Diagonals Properties
Symmetries
Trapezium
One pair of opposite sides is parallel ($AB \parallel CD$).
Diagonals divide each other proportionally.
Isosceles trapezium has line symmetry.
Parallelogram
Both pairs of opposite sides are parallel ($AB \parallel CD, AD \parallel BC$).
Diagonals bisect each other.
$180^\circ$ rotational symmetry.
Rhombus
Parallelogram with all 4 sides equal.
Diagonals bisect each other at right angles ($90^\circ$).
2 lines of symmetry (along diagonals).
Rectangle
Parallelogram with each angle $= 90^\circ$.
Diagonals are equal in length and bisect each other.
2 lines of symmetry (joining midpoints).
Square
Regular quadrilateral: 4 equal sides & 4 right angles.
Diagonals are equal and bisect each other at $90^\circ$.
4 lines of symmetry.
Kite
Two distinct pairs of adjacent sides are equal ($AB=AD, CB=CD$).
Diagonals intersect at $90^\circ$; one diagonal bisects the other.
1 line of symmetry.
3. Master Parallelogram Theorems
Core Euclidean Theorems
Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles ($\triangle ABC \cong \triangle CDA$).
Theorem 2: In a parallelogram, opposite sides are equal ($AB = CD$ and $BC = DA$).
Theorem 3: In a parallelogram, opposite angles are equal ($\angle A = \angle C$ and $\angle B = \angle D$).
Theorem 4: The diagonals of a parallelogram bisect each other ($OA = OC$ and $OB = OD$).
Theorem 5 (Sufficient Condition): A quadrilateral is a parallelogram if one pair of opposite sides is both equal and parallel ($AB = CD$ and $AB \parallel CD$).
4. The Mid-Point Theorem of a Triangle (Master Theorem)
This is the central pillar of class 9 geometry, serving as the bridge between triangles and quadrilaterals.
The Mid-Point Theorem
Statement: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
$$\text{In } \triangle ABC, \text{ if } D \text{ is midpoint of } AB \text{ and } E \text{ is midpoint of } AC:$$
$$DE \parallel BC \quad \text{and} \quad DE = \frac{1}{2} BC$$
Figure 12.1: Geometric proof construction of the Mid-Point Theorem ($DE \parallel BC, DE = \frac{1}{2}BC$).
Full Deductive Proof of the Mid-Point Theorem
Given: In $\triangle ABC$, $D$ is the midpoint of $AB$ ($AD = DB$) and $E$ is the midpoint of $AC$ ($AE = EC$).
To Prove: $DE \parallel BC$ and $DE = \frac{1}{2} BC$.
Construction: Extend $DE$ to point $F$ such that $EF = DE$. Join $CF$ such that $CF \parallel BA$.
Congruence of Triangles: In $\triangle ADE$ and $\triangle CFE$:
Forming Parallelogram $DBCF$: Since $AD = DB$ (given) and $AD = CF$, we have $BD = CF$.
Also, $BD \parallel CF$ (since $AB \parallel CF$).
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
Thus, $DBCF$ is a parallelogram!
Final Deduction: In parallelogram $DBCF$, opposite side $DF \parallel BC \implies DE \parallel BC$.
Also, $DF = BC \implies DE + EF = BC \implies 2DE = BC \implies DE = \frac{1}{2} BC$. Q.E.D.
$$\text{Converse of Mid-Point Theorem: } \text{The line drawn through the midpoint of one side of a triangle,}$$
$$\text{parallel to another side, bisects the third side!}$$
5. Varignon's Theorem (Midpoint Parallelogram)
Pierre Varignon established a stunning geometric theorem in 1731 that holds true for any quadrilateral whatsoever:
Varignon's Master Theorem
Statement: The figure formed by joining the midpoints of the adjacent sides of ANY quadrilateral (convex, concave, or self-intersecting) consecutively is ALWAYS a parallelogram.
Proof using Mid-Point Theorem: Let $P, Q, R, S$ be midpoints of sides $AB, BC, CD, DA$ of quadrilateral $ABCD$.
In $\triangle ABC$, $P$ and $Q$ are midpoints of $AB$ and $BC \implies PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
In $\triangle ADC$, $S$ and $R$ are midpoints of $AD$ and $CD \implies SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Therefore, $PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides is both equal and parallel, $PQRS$ is definitively a parallelogram!
Original Quadrilateral $ABCD$
Varignon Midpoint Parallelogram $PQRS$
Reason
Any Arbitrary Quadrilateral
Parallelogram
Always true by Mid-Point Theorem.
Rectangle
Rhombus
Original diagonals are equal $\implies$ adjacent sides of $PQRS$ are equal.
Rhombus
Rectangle
Original diagonals are perpendicular $\implies$ angles of $PQRS$ are $90^\circ$.
Square
Square
Diagonals are both equal AND perpendicular.
Kite
Rectangle
Diagonals of a kite are perpendicular ($AC \perp BD$).
6. Centroid, Bimedians & Tiling the Plane
Bimedians: The line segments connecting the midpoints of opposite sides of a quadrilateral ($PR$ and $QS$).
Centroid of a Quadrilateral: The bimedians of any quadrilateral bisect each other at a single common point, which is the centroid of the four vertices!
Tiling the Plane (Tessellations): Because the four interior angles of any quadrilateral sum to $360^\circ$, identical copies of any quadrilateral (even non-convex) can tile the entire 2D plane infinitely without any gaps or overlaps.
Example 1: Finding Angles in a Parallelogram NCERT
Problem: In a parallelogram $ABCD$, adjacent angles are in the ratio $2 : 3$. Find the measures of all four angles of the parallelogram.
Solution:
Let adjacent angles $\angle A$ and $\angle B$ be $2x$ and $3x$.
Since consecutive interior angles between parallel lines sum to $180^\circ$:
$$\angle A + \angle B = 180^\circ \implies 2x + 3x = 180^\circ \implies 5x = 180^\circ \implies x = 36^\circ$$
Therefore:
$$\angle A = 2 \times 36^\circ = 72^\circ$$
$$\angle B = 3 \times 36^\circ = 108^\circ$$
Since opposite angles in a parallelogram are equal:
$$\angle C = \angle A = 72^\circ, \quad \angle D = \angle B = 108^\circ$$
Angles: $72^\circ, 108^\circ, 72^\circ, 108^\circ$.
Example 2: Application of Mid-Point Theorem HOTS
Problem: In $\triangle ABC$, $D, E, F$ are the midpoints of sides $AB, BC, CA$ respectively. If the perimeter of $\triangle ABC$ is $38\text{ cm}$, find the perimeter of $\triangle DEF$.
Solution:
By the Mid-Point Theorem:
$$DE = \frac{1}{2} AC, \quad EF = \frac{1}{2} AB, \quad FD = \frac{1}{2} BC$$
Perimeter of $\triangle DEF = DE + EF + FD$:
$$\text{Perimeter}(\triangle DEF) = \frac{1}{2}AC + \frac{1}{2}AB + \frac{1}{2}BC = \frac{1}{2}(AB + BC + CA)$$
$$\text{Perimeter}(\triangle DEF) = \frac{1}{2} \times 38\text{ cm} = 19\text{ cm}$$
Answer: The perimeter of the midpoint triangle is exactly half the original perimeter, i.e., $19\text{ cm}$.
8. Chapter Summary & Quick Revision Checklist
Key Revision Points
Sum of interior angles of any quadrilateral is $360^\circ$.
In a parallelogram: opposite sides are equal, opposite angles are equal, diagonals bisect each other.
Mid-Point Theorem: $DE \parallel BC$ and $DE = \frac{1}{2}BC$.
Varignon's Theorem: Joining consecutive midpoints of ANY quadrilateral always forms a parallelogram.
Varignon parallelogram of a rectangle is a rhombus; of a rhombus is a rectangle.
VARDAAN COMET