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 AB= BC = CD= DA and AB ||CD , AD||BC ....................(1) Also diagonals of a parallelogram bisect each other. AO = OC and BO = OD ...................(2) In BOC and COD BO = DO [ From Equation (2)] BC = CD [ From Equation (1)] OC = OC [ Common ] Therfore BOC DOC [SSS Criteria of Congruence] [ By CPCT ] ............(3) But [Linear Pair] [Using Eq 3]
Similarly we can prove that 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 BD To Prove: ABCD is a rhombus. Proof: In AOB and COD AO = OC [ From Equation (1)] BO = OD [ From Equation (1)] AOB = COD [ Each as AC BD] Therefore AOB COD [ SAS Criteria of Congurence] BAO = DCO [CPCT] But they are alternate angles When AB and CD are straight lines and AC is the transversal AB || CD Similarly we can Prove that AD || BC As both pairs of opposite sides are equal Therefore ABCD is a ||gm In AOB and COB AO = OC [ From Equation (1)] BO = OB [ Common] AOB = COB [ Each as AC BD] Therefore AOB COB [ 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 find | |
Solution: As ABCD is a rhombus and we know diagonals of rhombus bisect at right angle. In [Exterior angle = Sum of Interior Opposite Angles] As every rhombus is a parallelogram, Therefore CD || AB [Alternate Interior Angles when CD || AB and BD is the transversal] In the figure [Linear pairs when ACE is a ray and DC stands on it ] In [ Angles opposite to equal sides are equal and AD = DC sides of a rhombus] Hence |
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