In this video we will learn about the different properties of quadrilaterals.
We have learnt that any four sided closed figure is called a Quadrilateral. Each quadrilateral has two parameters, angles and sides.
Based on the sides and angles there are certain properties of quadrilaterals. We are discussing them below:
Properties based on Length of Sides
- The Mid-Point Theorem
Joining mid-points of all sides of a quadrilateral forms a parallelogram. We take a quadrilateral ABCD and join their mid-points to form a quadrilateral EFGH. We draw a diagonal DB dividing the quadrilateral into ΔADB and ΔDBC. The Midpoint Theorem states that the segment joining two sides of a triangle at the mid-points of those sides is parallel to the third side and is half the length of the third side. Thus, HE || DB and HE = ½ DB.
Likewise, GF || DB and GF= ½ DB
Similarly when we draw the other diagonal AC, EF and HG || AC and EF = HG = ½ AC.
So we see that HE || GF and EF || HG. Thus, they form a parallelogram EFGH.
- Joining the Mid-Points of a Rhombus forms a Rectangle and vice versa.
We know that both rhombus and rectangle are different types of parallelograms. So, when we join the mid points of the four sides of a rhombus we get a rectangle. And likewise when we join the mid-points of a rectangle, we get a rhombus.
Properties based on angles
- Bisectors of Alternate Angles on Parallel lines form a Rectangle.
AB and CD are two parallel lines and EF is a transversal intersecting them. By the property of alternate angles, ∠CFE and ∠BEF will be equal to each other. By the same property, ∠AEF and ∠DFE will also be equal to each other. We know that angles on a straight line are supplementary. Thus, ∠CFE + ∠EFD = 180° and ∠AEF + ∠BEF = 180°. Each of these angles is bisected. The bisectors of ∠AEF and ∠CFE meet at a point G whereas the bisectors of ∠FEB and ∠EFD meet at a point H, thus forming GEF and FHE respectively. On bisecting these angles, the sum of the angles on the straight line will also be halved. This means that ∠GFH and ∠GEF = 90°. Similarly, ∠EGF and ∠EHF are also equal to 90°. Thus the figure EFGH is a rectangle.
- Bisectors of Interior Angles of a Parallelogram enclose a Rectangle.
To illustrate this we take a parallelogram ABCD, and bisect its interior angles ∠DAB, ∠ADC, ∠ABC and ∠BCD. We know that adjacent angles of a parallelogram are supplementary. Thus, ∠DAB and ∠ADC add up to 180°. The bisectors of ∠DAB and ∠ADC meet at a point E, thus forming ΔADE. On bisecting them, the sum of their angles will be halved, i.e. 90°. The sum of angles of a triangle is 180°, so the ∠DEA will be 90°. This angle is vertically opposite to the interior angle of the figure enclosed by the bisectors. Similarly, the other interior angles of the figure will also be 90°. And as we have studied earlier, a quadrilateral whose 90° is a rectangle. Thus, the figure enclosed by the Bisector of interior angles of a parallelogram is a rectangle.