The Diagonal of a Rhombus are Perpendicular to Each Other
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
Also diagonals of a parallelogram bisect each other.
In BO = DO [ From Equation (2)] BC = CD [ From Equation (1)] OC = OC [ Common ] Therfore But
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 To Prove: ABCD is a rhombus. Proof: In AO = OC [ From Equation (1)] BO = OD [ From Equation (1)] Therefore
But they are alternate angles When AB and CD are straight lines and AC is the transversal
Similarly we can Prove that AD || BC As both pairs of opposite sides are equal Therefore ABCD is a ||gm In AO = OC [ From Equation (1)] BO = OB [ Common] Therefore 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 | |
Solution: As ABCD is a rhombus and we know diagonals of rhombus bisect at right angle. In
As every rhombus is a parallelogram, Therefore CD || AB
In the figure
In
Hence |  |
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