if diagonals of a parallelogram bisect each other at 90 degrees prove it is a rhombus
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Answer:
The exercise above showed that each diagonal of a rhombus dissects the rhombus into two congruent triangles that are reflections of each other in the diagonal,
triangleABC equiv triangleADC and triangleBAD equiv triangleBCD.
Thus the diagonals of a rhombus are axes of symmetry.
The following property shows that these two axes are perpendicular.
Third property of a rhombus − The diagonals are perpendicular
The proof given here uses the theorem about the axis of symmetry of an isosceles triangle proven at the start of this module. Two other proofs are outlined as exercises.
Theorem
The diagonals of a rhombus are perpendicular.
Proof
Let ABCD be a rhombus,
with diagonals meeting at M.
To prove that AC ⊥ BD.
By the previous theorem, AM is the angle bisector of triangleDAB.
Hence AM ⊥ BD, because A is the apex of the isosceles triangle ABD,
The diagonals also bisect each other because a rhombus is a parallelogram, so we usually state the property as
‘The diagonals of a rhombus bisect each other at right angles.’
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EXERCISE 5
a
Use congruence to prove this property.
b
Use angle-chasing to prove this property.
We now turn to tests for a quadrilateral to be a rhombus. This is a matter of establishing that a property, or a combination of properties, gives us enough information for us to conclude that such a quadrilateral is a rhombus.
First test for a rhombus − A parallelogram with two adjacent sides equal
We have proved that the opposite sides of a parallelogram are equal, so if two adjacent sides are equal, then all four sides are equal and it is a rhombus.
Theorem
If two adjacent sides of a parallelogram are equal, then it is a rhombus.
This test is often taken as the definition of a rhombus.
Second test for a rhombus − A quadrilateral whose diagonals bisect each other
at right angles
Theorem
A quadrilateral whose diagonals bisect each other at right angles is a rhombus.
Proof
Let ABCD be a quadrilateral whose diagonals bisect
each other at right angles at M.
We prove that DA = AB. It follows similarly that
AB = BC and BC = CD
triangleAMB equiv triangleAMD (SAS)
So AB = AD and by the first test above ABCD is a rhombus.
A quadrilateral whose diagonals bisect each other is a parallelogram, so this test is often stated as
‘If the diagonals of a parallelogram are perpendicular, then it is a rhombus.’
This test gives us another construction of a rhombus.
Construct two perpendicular lines intersecting
at M.
Draw two circles with centre M and different radii.
Join the points where alternate circles cut the lines.
This figure is a rhombus because its diagonals bisect each other at right angles.
Step-by-step explanation: