The Diagonal of a Rhombus are Perpendicular to Each Other


 
 
Concept Explanation
 

The Diagonal of a Rhombus are Perpendicular to Each Other

Theorem: The diagonal of a rhombus are perpendicular to each other

Given: A rhombus ABCD in which AC and BD are diagonals.

To Prove: AC and BC intersect perpendicularly.

Proof: As ABCD is a rhombus Therefore it is a parallelogram

Rightarrow AB= BC = CD= DA and AB ||CD , AD||BC ....................(1)

Also diagonals of a parallelogram bisect each other.

Rightarrow AO = OC and BO = OD              ...................(2)

In Delta BOC and Delta COD

  BO = DO       [ From Equation (2)]

  BC = CD       [ From Equation (1)]

 OC = OC       [ Common ]

Therfore Delta BOC cong Delta DOC     [SSS Criteria of Congruence]

  Rightarrow angle COB = angle COD     [ By CPCT ]    ............(3)

But angle COB + angle COD = 180^0               [Linear Pair]

angle COB + angle COB = 180^0                     [Using Eq 3]

  2 angle COB = 180^0Rightarrow angle COB = 90^0

Rightarrow angle COB = angle COD = 90^0

Similarly we can prove that angle AOB = angle AOD = 90^0

angle AOB = angle AOD = angle COB = angle COD =90^0

Hence Proved.

Theorem: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Given: A quadrilateral ABCD such that AC and CD bisect at Right angle.

           AO = OC   and BO = OD             ....................(1)

           AC perp BD                  

To Prove: ABCD is a rhombus.

Proof: In Delta AOB and DeltaCOD

         AO = OC      [ From Equation (1)]

         BO = OD      [ From Equation (1)]

   angle AOB = angleCOD  [ Each 90^0 as AC perp BD]

Therefore Delta AOB cong DeltaCOD     [ SAS Criteria of Congurence]

Rightarrow angle BAO = angleDCO             [CPCT]

But they are alternate angles When AB and CD are straight lines and AC is the transversal

Rightarrow AB || CD

Similarly we can Prove that AD || BC

As both pairs of opposite sides are equal Therefore ABCD is a ||gm

In Delta AOB and DeltaCOB

         AO = OC      [ From Equation (1)]

         BO = OB      [ Common]

   angle AOB = angleCOB  [ Each 90^0 as AC perp BD]

Therefore Delta AOB cong DeltaCOB     [ SAS Criteria of Congurence]

     AB = BC     [CPCT]

As ABCD is a IIgm and the adjacent sides are equal AB = BC Therefore ABCD is a rhombus

Illustration: ABCD is a rhombus in which diagonal AC is produced to E . If angle DCE = 138^0 find angle DOC, angle ABD; and ; angle DAC

Solution:  As ABCD is a rhombus and we know diagonals of rhombus bisect at right angle.

therefore angle DOC= 90^0

In Delta DOC 

angle CDO +angle DOC = angle DCE     [Exterior angle = Sum of Interior Opposite Angles]

angle CDO +90^0 = 138^0Rightarrow angle CDO = 138^0- 90^0= 48^0

As every rhombus is a parallelogram, Therefore CD || AB

Rightarrow angle ABD =angle CDO     [Alternate Interior Angles when CD || AB and BD is the transversal]

Rightarrow angle ABD =48^0

In the figure

angle ACD+ angle DCE = 180^0   [Linear pairs when ACE is a ray and DC stands on it ]

angle ACD+ 138^0 = 180^0Rightarrow angle ACD = 180 - 138= 42^0

In Delta ADC

angle DAC= angle ACD    [ Angles opposite to equal sides are equal and AD = DC sides of a rhombus]

angle DAC= 42^0

Hence angle DOC=90^0, angle ABD= 48^0; and ; angle DAC= 42^0

 
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