special parallelogram

Special Parallelogram

So far in this chapter, we have studied about parallelograms and their properties. In this section we will study some special parallelogram. As panther belongs to the cat family, rectangles, rhombus and square belong to the family of parallelograms. These three special parallelograms are defined bellows: 1. Rectangle A rectangle is a parallelogram with one of its angles a right angle. We know that in a parallelogram, the adjacent angles are supplementary and opposite angles are equal; therefore, a rectangle is special parallelogram each of whose angles is a right angle. Property 1: Each of the angles of a rectangle is a right angle Given ABCD is a rectangle in which A  90o . To prove: A  B  C  D  90o



Proof: Since ABCD is a rectangle, therefore it is a parallelogram



or AD parallel to BC Since AD parallel to BC and AB is the transversal.

A  B  180o



(Consecutive interior angles are supplementary) ( Since A  90o )



or 90o  B  180o or B  180o  90o  90o



We know that opposite angles of a parallelogram are equal Therefore D  90o and C  90o (A rectangle is a parallelogram)



Hence A  B  C  D  90o Property 2: The diagonals of a rectangle are equal. Given: ABCD is a rectangle with AC and BD as its diagonals. To prove: AC=BD



Proof: By definition, we know that a rectangle is a parallelogram with one of its angles  90o .



Let A  90o Since AD parallel to BC and AB is the transversal (Since Opposite sides of a parallelogram are parallel to each other) Therefore A  B  180o or 90o  B  180o or B  180o  90o  90o In Triangle ABD and Triangle BAC , we find AD=BC

A  B  90o



(opposite sides of a parallelogram are equal) (Proved above) (Common to both the triangles)



AB=AB



Therefore BY SAS criterion of congruency

Triangle ABD  Triangle BAC Therefore AC  BD



(cpctc)



Therefore Diagonals of a rectangle are of equal length. Property 3: (Converse of property 2). If two diagonals of a parallelogram are equal, it is a rectangle. Given: A parallelogram ABCD in which AC=BD.



To prove: ABCD is a rectangle Proof: In Triangle ABC and Triangle DCB AB=DC BC=BC AC=BD (Opposite sides of a parallelogram are equal) (Common to the triangles) (Given)



Therefore By SSS criterion of congruence,

Triangle ABC  Triangle DCB



Thus, ABC  DCB



… (1) (cpctc)



Now, AB parallel to DC and BC intersects, them Therefore ABC  DCB  180o … (2) Consecutive interior angles are supplementary



or ABC  ABC  180o or 2ABC  180o



or ABC 



180o  90o 2



Thus, ABC  DCB  90o So, we conclude that ABCD is a parallelogram one of whose angle is 90o . Hence ABCD is a rectangle. 2. Rhombus A rhombus is a parallelogram with a pair of its adjacent sides equal. From our knowledge of parallelograms, we already know that in a parallelogram, both pairs of opposite sides are equal. Hence, a rhombus is a parallelogram, whose all four sides are equal. Property 1: Each of the four sides of a rhombus is of the same length. Given: In figure 7-28, ABCD is a rhombus such that AB=BC To prove: AB=BC=CD=DA Proof: Since ABCD is a rhombus, this implies that ABCD is a parallelogram (By definition) Thus AB=CD and BC=AD (Opposite sides of parallelogram are equal) Now AB=BC (Given)

Therefore AB  BC  CD  DA



Hence, all the four sides of a rhombus are equal.



Property 2: The diagonals of a rhombus are perpendicular to each other. Given: ABCD is a rhombus. Diagonals AC and BD intersect each other at O. To prove: AC and BP are perpendicular AOB  BOC  COD  DOA  90o to each other, that is



Proof: Since ABCD is a rhombus

Therefore AB  BC  CD  DA



… (1) By definition



Since, the diagonals of a parallelogram bisect each other,

Therefore OA  OC and OB  OD



… (2)



In Triangle AOB and Triangle BOC , we have AB=BC OA=OC and OB=OB (by 1) (by 2) (Common to both the triangles)



Therefore Triangle AOB  Triangle BOC (By SSS criterion of congruency) Therefore AOB  COB (cpctc)



But AOB  COB  180o (They form a linear pair) or COB  COB  180o or 2COB  180o or COB  180o  90o 2 ( Since AOB  COB )



or Therefore AOB  90o So, we conclude that AC  BD .



Hence diagonals of a rhombus are perpendicular to each other. Property 3: (Converse of property 2) If the diagonals of a parallelogram are perpendicular, then it’s a rhombus Given: A parallelogram ABCD in which AC  BD . To prove: Parallelogram ABCD is a rhombus Proof: Let AC and BD intersect at O. Then OA=OC and OB=OD (Diagonals of a parallelogram bisect each other)



In Triangle AOB and Triangle COB , we find OA=OC

AOB  COB  90o OB=OB



(shown above) ( Since AC  BD ) (Common side of the triangles) (By SAS criterion of congruency)



Therefore Triangle AOB  Triangle COB



Hence, AB=BC



(cpctc)



… (1)



Since, ABC is a parallelogram

Therefore AB  DC and BC=AD



… (2)



From (1) and (2), we get AB=BC=AD=BC Hence, parallelogram ABCD is a rhombus. 3. Square A square is a rectangle with a pair of consecutive sides equal. Since in a rectangle, the opposite sides are equal, therefore, in a square opposite sides are also equal. Since in a square, consecutive sides are equal, therefore, sides



opposite to consecutive sides will also be equal. Hence, in a square, all the four sides are equal and angle is a right angle. Property 1: Each of the angles of a square is a right angle and each of the four sides is of the same length. Given: A square ABCD such that AB=BC To prove: AB=BC=CD=DA and A  B  C  C  90o



Proof: Since ABCD is a square. Therefore ABCD is a rectangle (Since square is a special rectangle) So, A  B  C  C  90o Again, since ABCD is a square Therefore ABCD is a parallelogram such that AB=BC So, A  B  C  C  90o Property 2: Diagonals of a square are equal and perpendicular to each other. Given: Square ABCD To prove: AC=BD and AC  BD



Proof: In Triangle ADB and Triangle BCA , we find AD=BC

BAD  ABC



(All sides of a square are equal) (Each is a right angle) (Common to both the triangles)



and AB=BA



Therefore By SAS criterion of congruence,

Triangle ADB  Triangle BCA



Thus AC=BD



(cpctc)



Now, in Triangle AOB and Triangle AOD , we have OB=OD AB=AD AO=AO (Diagonals of a parallelogram bisect each other) (sides of a square are equal) (Common to both the triangles) (By SSS criterion of congruence)



So, Triangle AOB  Triangle AOD Thus, AOB  AOD (cpctc)



Now AOB  AOD  180o (these form a linear pair) Therefore AOD  AOD  180o

2AOD  180o



( Since AOB  AOD )



180o AOD   90o 2 and AOB  90o Hence, AO  BD  AC  BD Therefore, AC=BD and AC  BD Property 3: If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a square. Given: Parallelogram ABCD in which AC=BD and AC  BD To prove: ABCD is a square



Proof: In Triangle AOB and Triangle AED , we have OB=OD (diagonals of a parallelogram bisect each other) OA=OA (common to both the triangles)



AOB  AOD



(Each = 90o as AC  BD )



Therefore Triangle AOB  Triangle AOD (By SAS criterion of congruence)



So, AB=AD



(cpctc)



… (1)



Since, in a parallelogram, opposite sides are equal Therefore AB=CD and AD=BC So, AB=BC=CD=AD [By using (1) and (2)] Now in Triangle ABD and Triangle BAC , we find AD=BC AC=BD AB=AC (by 2) (Given) (Common to both the triangles) (By SSS congruence criterion) … (2)



Therefore Triangle ABD  Triangle BAC



So, DAB  CBA (cpctc) But, DAB  CBA  180o (these form consecutive interior angles) Therefore DAB  CBA  90o Since, opposite angles of a parallelogram are equal

DAB  BCD  90o



And CBA  ADC  90o Thus, AB=BC=CD=DA and A  B  C  D  90o Hence, ABCD is a square.