Quadrilaterals

Hello students, welcome to today's mathematics lesson. I am so happy to be here with you to learn about quadrilaterals, which is Chapter 8 of your NCERT textbook. Today we are going to explore some very interesting properties of quadrilaterals, especially parallelograms, and we will also learn about a very useful theorem related to triangles. So let's begin our journey into the world of geometry.

You have already studied quadrilaterals and their types in Class VIII. Let me remind you what a quadrilateral is. A quadrilateral is a four-sided figure - it has four sides, four angles, and four vertices. Now, in this chapter, we are going to study a special type of quadrilateral called a parallelogram. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. So if you have a quadrilateral ABCD, it is a parallelogram if side AB is parallel to side CD, and side AD is parallel to side BC. This is the basic definition that you must remember.

Now, let us perform an activity to understand a very important property of parallelograms. Imagine you have a sheet of paper and you cut out a parallelogram shape from it. Now, take a pair of scissors and cut this parallelogram along one of its diagonals - that is, from one corner to the opposite corner. What do you get? You get two triangles. Now, take one triangle and place it over the other triangle. You may need to turn one around to match them properly. What do you observe? You will see that the two triangles are exactly the same in size and shape. In geometry, we say they are congruent to each other. This is a very interesting observation. Let me tell you something wonderful - this happens with every parallelogram, not just the one you cut. If you repeat this activity with different parallelograms, each time you will find that each diagonal divides the parallelogram into two congruent triangles. This is such an important result that we prove it as a theorem.

So here is Theorem 8.1: A diagonal of a parallelogram divides it into two congruent triangles.

Let us prove this theorem. Let ABCD be a parallelogram and AC be a diagonal. The diagonal AC divides the parallelogram ABCD into two triangles, namely triangle ABC and triangle CDA. We need to prove that these two triangles are congruent.

Look at triangle ABC and triangle CDA. In these triangles, note that BC is parallel to AD and AC is a transversal cutting these parallel lines. So, angle BCA equals angle DAC because they are a pair of alternate interior angles. Also, AB is parallel to DC and AC is a transversal. So, angle BAC equals angle DCA, again because they are alternate interior angles. And of course, AC is common to both triangles - it is the same side in both triangles. So we have two angles and the included side equal in both triangles. By the ASA congruence rule, triangle ABC is congruent to triangle CDA. Therefore, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA. This is exactly what we wanted to prove.

Now, let us think about what this result tells us. If two triangles are congruent, then all their corresponding parts are equal. That means the sides that correspond to each other are equal in length. So, from triangle ABC being congruent to triangle CDA, we can say that side AB corresponds to side CD, and side AD corresponds to side BC. Therefore, AB equals DC and AD equals BC. This gives us another very important property of parallelograms.

Here is Theorem 8.2: In a parallelogram, opposite sides are equal.

So if ABCD is a parallelogram, then AB equals DC and AD equals BC. This is a property that we use very often in solving geometry problems. Remember, this works in both directions. If you know that a quadrilateral is a parallelogram, then you can conclude that its opposite sides are equal. But the question is - what if we have a quadrilateral in which the opposite sides are equal? Will it necessarily be a parallelogram? The answer is yes, and this is the converse of Theorem 8.2.

Let me explain what a converse means. In mathematics, when we have a theorem that says "if A, then B", the converse is "if B, then A". So Theorem 8.2 says: If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. The converse would be: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. And this is indeed true.

Here is Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Let us reason this out. Suppose we have a quadrilateral ABCD in which AB equals CD and AD equals BC. Draw the diagonal AC. Now look at triangle ABC and triangle CDA. In these triangles, we have AB equals CD (given), AD equals BC (given), and AC is common. So all three sides of triangle ABC are equal to the corresponding three sides of triangle CDA. By the SSS congruence rule, triangle ABC is congruent to triangle CDA. Therefore, angle BAC equals angle DCA, and angle BCA equals angle DAC. Now, if angle BAC equals angle DCA, these are alternate interior angles when lines AB and CD are cut by transversal AC. This means AB is parallel to CD. Similarly, if angle BCA equals angle DAC, then lines BC and AD are parallel. So both pairs of opposite sides are parallel, which means ABCD is a parallelogram. That proves Theorem 8.3.

Now, let us think about the angles of a parallelogram. Draw a parallelogram and measure its angles using a protractor. What do you observe? You will find that each pair of opposite angles is equal. If you repeat this with some more parallelograms, you will get the same result. So we have another property.

Theorem 8.4: In a parallelogram, opposite angles are equal.

Now, is the converse of this also true? That is, if in a quadrilateral each pair of opposite angles is equal, will it be a parallelogram? Yes, this is also true. Using the angle sum property of a quadrilateral and the results of parallel lines intersected by a transversal, we can prove this.

Theorem 8.5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Now we come to yet another very important property of parallelograms - the property related to their diagonals. Draw a parallelogram ABCD and draw both its diagonals. They intersect each other at a point. Let us call this point O. Now measure the lengths of OA, OB, OC, and OD. What do you observe? You will find that OA equals OC and OB equals OD. In other words, the point O is the mid-point of both the diagonals. If you repeat this activity with different parallelograms, you will find that this is always true. The diagonals of a parallelogram always bisect each other.

Theorem 8.6: The diagonals of a parallelogram bisect each other.

Now, what would happen if in a quadrilateral the diagonals bisect each other? Will it be a parallelogram? Yes, this is true as well. This is the converse of Theorem 8.6.

Theorem 8.7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Let me prove this for you. In the figure, it is given that OA equals OC and OB equals OD. So consider triangle AOB and triangle COD. In these triangles, OA equals OC, OB equals OD, and angle AOB equals angle COD because they are vertically opposite angles. So by the SAS congruence rule, triangle AOB is congruent to triangle COD. Therefore, angle ABO equals angle CDO, which means these are alternate interior angles when lines AB and CD are cut by transversal BD. So AB is parallel to CD. Similarly, we can prove that BC is parallel to AD. Therefore, ABCD is a parallelogram.

So students, let me summarize what we have learned so far about parallelograms. We have learned that in a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other. These are the three main properties of a parallelogram. And we have also learned the converses of these properties, which tell us that if any of these conditions are satisfied in a quadrilateral, then it must be a parallelogram.

Now, let us look at some examples to understand how these properties are used in solving problems.

Example 1: Show that each angle of a rectangle is a right angle.

First, let us recall what a rectangle is. A rectangle is a parallelogram in which one angle is a right angle. So if ABCD is a rectangle, we know that it is a parallelogram, and let us say angle A equals 90 degrees. We need to show that angles B, C, and D are also 90 degrees each.

Since ABCD is a rectangle, it is a parallelogram. So AD is parallel to BC, and AB is a transversal cutting these parallel lines. Therefore, angle A and angle B are interior angles on the same side of the transversal, so their sum is 180 degrees. That is, angle A plus angle B equals 180 degrees. But angle A is 90 degrees, so angle B equals 180 degrees minus 90 degrees, which is 90 degrees.

Now, in a parallelogram, opposite angles are equal. So angle C equals angle A, which is 90 degrees, and angle D equals angle B, which is also 90 degrees. Therefore, each of the angles of a rectangle is a right angle. This is what we wanted to prove.

Example 2: Show that the diagonals of a rhombus are perpendicular to each other.

First, let us understand what a rhombus is. A rhombus is a parallelogram in which all four sides are equal. So in rhombus ABCD, we have AB equals BC equals CD equals DA.

Now, we need to prove that the diagonals of a rhombus are perpendicular to each other, that is, they intersect at right angles.

Consider the rhombus ABCD. We know that in a rhombus, all sides are equal. Now, look at triangles AOD and COD. In these triangles, OA equals OC because the diagonals of a parallelogram bisect each other - this is Theorem 8.6 that we learned earlier. OD is common to both triangles. And AD equals CD because all sides of a rhombus are equal. So we have all three sides of triangle AOD equal to the corresponding three sides of triangle COD. By the SSS congruence rule, triangle AOD is congruent to triangle COD. Therefore, angle AOD equals angle COD, because these are corresponding parts of congruent triangles.

But angle AOD and angle COD form a linear pair, which means their sum is 180 degrees. So twice angle AOD equals 180 degrees, which gives us angle AOD equals 90 degrees. Therefore, the diagonals of a rhombus are perpendicular to each other. This is what we wanted to prove.

Example 3: This is a more complex example. ABC is an isosceles triangle in which AB equals AC. AD bisects the exterior angle PAC and CD is parallel to AB. We need to show that angle DAC equals angle BCA, and that ABCD is a parallelogram.

Let us solve this step by step. First, we are given that triangle ABC is isosceles with AB equals AC. In an isosceles triangle, the angles opposite to equal sides are equal. So angle ABC equals angle ACB.

Now, consider the exterior angle PAC of triangle ABC. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. So angle PAC equals angle ABC plus angle ACB. But since angle ABC equals angle ACB, we can write angle PAC equals 2 times angle ACB.

Now, AD bisects angle PAC. So angle PAC equals 2 times angle DAC.

From these two equations, we get 2 times angle DAC equals 2 times angle ACB, which gives us angle DAC equals angle ACB, or in other words, angle DAC equals angle BCA. This proves the first part.

For the second part, we need to show that ABCD is a parallelogram. We have just proved that angle DAC equals angle BCA. These equal angles form a pair of alternate interior angles when line segments BC and AD are intersected by transversal AC. So BC is parallel to AD.

Also, it is given that CD is parallel to AB.

Now, in quadrilateral ABCD, we have both pairs of opposite sides parallel - AB is parallel to CD, and BC is parallel to AD. Therefore, ABCD is a parallelogram. This is what we wanted to prove.

Example 4: Two parallel lines l and m are intersected by a transversal p. We need to show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Let me explain this problem. We have two parallel lines l and m, and a transversal p that cuts these parallel lines at points A and C respectively. Now, we take the bisectors of angle PAC and angle ACQ - they intersect at point B. Similarly, the bisectors of angle ACR and angle SAC intersect at point D. We need to show that quadrilateral ABCD is a rectangle.

First, let us show that ABCD is a parallelogram. Since lines l and m are parallel and p is a transversal, angle PAC equals angle ACR because they are alternate interior angles. Therefore, half of angle PAC equals half of angle ACR, which means angle BAC equals angle ACD. These are alternate interior angles for lines AB and DC with AC as transversal, and they are equal. So AB is parallel to DC.

Similarly, we can show that BC is parallel to AD. Therefore, quadrilateral ABCD is a parallelogram.

Now, we need to show that one angle of this parallelogram is 90 degrees. Consider angle PAC and angle CAS. They form a linear pair, so their sum is 180 degrees. Therefore, half of angle PAC plus half of angle CAS equals half of 180 degrees, which is 90 degrees. But half of angle PAC is angle BAC, and half of angle CAS is angle CAD. So angle BAC plus angle CAD equals 90 degrees, which means angle BAD equals 90 degrees.

So we have a parallelogram in which one angle is 90 degrees. Therefore, ABCD is a rectangle. This is what we wanted to prove.

Example 5: Show that the bisectors of angles of a parallelogram form a rectangle.

This is a very interesting example. Let P, Q, R, and S be the points of intersection of the bisectors of angles A and B, B and C, C and D, and D and A respectively of parallelogram ABCD.

We need to show that PQRS is a rectangle.

Consider triangle ASD. Since DS bisects angle D and AS bisects angle A, we have angle DAS equals half of angle A, and angle ADS equals half of angle D. So angle DAS plus angle ADS equals half of angle A plus half of angle D, which equals half of the sum of angle A and angle D.

Now, in parallelogram ABCD, angles A and D are interior angles on the same side of transversal AD. So their sum is 180 degrees. Therefore, angle DAS plus angle ADS equals half of 180 degrees, which is 90 degrees.

Now, in triangle ASD, the sum of all angles is 180 degrees. So angle DAS plus angle ADS plus angle DSA equals 180 degrees. But angle DAS plus angle ADS is 90 degrees, so angle DSA equals 180 degrees minus 90 degrees, which is 90 degrees.

Now, angle PSR is vertically opposite to angle DSA, so angle PSR is also 90 degrees.

Similarly, we can show that angle SPQ is 90 degrees, angle PQR is 90 degrees, and angle SRQ is 90 degrees.

So we have a quadrilateral PQRS in which all angles are 90 degrees. Therefore, PQRS is a parallelogram in which one angle is 90 degrees, which means it is a rectangle. This is what we wanted to prove.

Now students, we have completed the first section on the properties of a parallelogram. Let me quickly recap what we learned in this section. We learned that a diagonal of a parallelogram divides it into two congruent triangles. We learned that in a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other. We also learned the converses of these properties, which tell us that if any of these conditions are satisfied in a quadrilateral, then it is a parallelogram. We learned about rectangles and rhombuses as special types of parallelograms, and we proved that in a rectangle, all angles are right angles, and in a rhombus, the diagonals are perpendicular to each other. We also learned that the bisectors of angles of a parallelogram form a rectangle.

Now, let us move on to the second part of this chapter, which is about the mid-point theorem. This is a very useful theorem in geometry, and it relates to triangles.

You have studied many properties of a triangle as well as a quadrilateral. Now let us study yet another result which is related to the mid-point of sides of a triangle.

Perform the following activity. Draw a triangle ABC. Now, mark the mid-point E of side AB, and the mid-point F of side AC. Join the points E and F. Now, measure the length of EF and the length of BC. What do you observe? You will find that EF is exactly half of BC. Also, measure angle AEF and angle ABC. You will find that these angles are equal, which means EF is parallel to BC.

If you repeat this activity with some more triangles, you will get the same result. So we have discovered a very useful theorem.

Theorem 8.8: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

This theorem is also known as the mid-point theorem. It tells us that if we have a triangle and we join the mid-points of any two sides, the line segment joining them is parallel to the third side and is exactly half of it.

Let us prove this theorem. In the figure, E and F are the mid-points of AB and AC respectively. We need to prove that EF is parallel to BC and EF equals half of BC.

Consider the figure. Draw a line through C parallel to BA, and extend it to meet the line through E parallel to BC at a point D.

Now, look at triangle AEF and triangle CDF. Angle AEF equals angle CDF because they are corresponding angles as EF is parallel to BC and CD is drawn parallel to BA. Angle AFE equals angle CFD because they are also corresponding angles. And AF equals FC because F is the mid-point of AC. So by the ASA congruence rule, triangle AEF is congruent to triangle CDF.

Therefore, EF equals DF, and BE equals AE equals DC. Now, look at quadrilateral BCDE. We have BE equals DC, and BE is parallel to DC because BE is part of AB which is parallel to CD. So BCDE is a parallelogram. Therefore, EF is parallel to BC. Also, since BCDE is a parallelogram, ED equals BC. But EF equals half of ED, so EF equals half of BC. This proves Theorem 8.8.

Now, what is the converse of this theorem? The converse would be: If a line is drawn through the mid-point of one side of a triangle parallel to another side, then it bisects the third side. And this is also true.

Theorem 8.9: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

In the figure, E is the mid-point of AB, and a line through E parallel to BC meets AC at F. We need to prove that AF equals FC.

Draw a line through C parallel to BA, and let it meet the line through F parallel to BC at a point D. Now, we can prove that triangle AEF is congruent to triangle CDF, and hence AF equals FC. This proves Theorem 8.9.

Now, let us look at some examples related to the mid-point theorem.

Example 6: In triangle ABC, D, E, and F are respectively the mid-points of sides AB, BC, and CA. We need to show that triangle ABC is divided into four congruent triangles by joining D, E, and F.

This is a very interesting example. We have triangle ABC, and D, E, F are the mid-points of its sides. We join D to E, E to F, and F to D. This divides triangle ABC into four smaller triangles: triangle AFD, triangle BDE, triangle CEF, and triangle DEF. We need to show that all these four triangles are congruent to each other.

Since D and E are mid-points of sides AB and BC of triangle ABC, by Theorem 8.8, DE is parallel to AC. Similarly, DF is parallel to BC, and EF is parallel to AB.

Therefore, ADEF is a parallelogram, BDFE is a parallelogram, and DFCE is a parallelogram.

Now, consider parallelogram BDFE. DE is a diagonal of this parallelogram. So triangle BDE is congruent to triangle FED. Similarly, in parallelogram ADEF, DF is a diagonal, so triangle DAF is congruent to triangle FED. And in parallelogram DFCE, EF is a diagonal, so triangle EFC is congruent to triangle FED.

Therefore, all four triangles - triangle BDE, triangle DAF, triangle EFC, and triangle FED - are congruent to each other. This is what we wanted to prove.

Example 7: l, m, and n are three parallel lines intersected by transversals p and q such that l, m, and n cut off equal intercepts AB and BC on p. We need to show that l, m, and n cut off equal intercepts DE and EF on q also.

This is a problem that uses the mid-point theorem in a very clever way. We are given that AB equals BC, and we need to prove that DE equals EF.

Let us join A to F, and let this line intersect m at point G.

Now, consider the trapezium ACFD. It is divided into two triangles: triangle ACF and triangle AFD.

In triangle ACF, it is given that B is the mid-point of AC because AB equals BC. And BG is parallel to CF because m is parallel to n. So by Theorem 8.9, G is the mid-point of AF.

Now, in triangle AFD, we can apply the same argument. Since G is the mid-point of AF, GE is parallel to AD, and so by Theorem 8.9, E is the mid-point of DF. Therefore, DE equals EF.

In other words, l, m, and n cut off equal intercepts on q as well. This is what we wanted to prove.

Now students, we have completed the entire chapter. Let me summarize everything that we have learned in this chapter.

In this chapter, you have studied the following points:

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

In a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other.

The diagonals of a rectangle bisect each other and are equal, and vice-versa. This means that if a parallelogram has equal diagonals, it must be a rectangle.

The diagonals of a rhombus bisect each other at right angles, and vice-versa. This means that if a parallelogram has perpendicular diagonals, it must be a rhombus.

The diagonals of a square bisect each other at right angles and are equal, and vice-versa. This means that if a parallelogram has equal diagonals that are also perpendicular, it must be a square.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it. This is Theorem 8.8, which is also known as the mid-point theorem.

A line through the mid-point of a side of a triangle parallel to another side bisects the third side. This is Theorem 8.9, which is the converse of the mid-point theorem.

These are all the important concepts, definitions, and theorems from Chapter 8 on Quadrilaterals. I hope you have understood everything clearly. Remember, geometry is all about understanding properties and theorems, and then applying them to solve problems. So make sure you remember these properties well, as they will be very useful in your future studies and in solving problems.

Thank you for listening to this lesson. Keep practicing, and good luck with your studies!