Prerequisites:
Rhombus (definition)
Parallelogram (definition)
Alternate angles property (proof)
SSS congruency (proof)
Properties of parallelogram (proof)
Angle on a straight line (proof)
Proof:
Let ABCD be a rhombus with AC being one of its diagonal.
In \triangle ABC and \triangle ADC,
\qquad\quad AB = CD\qquad\qquad\qquad\qquad\qquad\text{(by definition of rhombus)}\\ \text{Also,}\;\;\: BC = AD\qquad\qquad\qquad\qquad\qquad\text{(by definition of rhombus)}\\ \qquad\quad AC = AC\qquad\qquad\qquad\qquad\qquad\:\!\text{(common)}
Hence, \triangle ABC\cong\triangle ADC by SSS rule.
Therefore, by CPCTC, \angle BAC = \angle DCA.
Hence by the alternate angles property, AB \parallel CD.
Similarly, as \angle BCA = \angle DAC, BC \parallel AD.
Therefore, ABCD is a parallelogram (see definition of parallelogram).
Now, let BD be another diagonal of ABCD, which intersects AC at E. Since ABCD is a parallelogram, therefore by the property of a parallelogram, AC and BD bisects each other.
In \triangle AED and \triangle CED,
\qquad\quad AD = CD\qquad\qquad\qquad\qquad\qquad\text{(by definition of rhombus)}\\ \text{Also,}\;\;\: AE = CE\qquad\qquad\qquad\qquad\qquad\:\!\text{(property of parallelogram)}\\ \qquad\quad DE = DE\qquad\qquad\qquad\qquad\qquad\text{(common)}
Hence, \triangle AED\cong\triangle CED by SSS rule.
Therefore, by CPCTC, \angle AED = \angle DEC.
Also, since AC is a straight line,
\therefore\quad\;\;\;\angle AED + \angle DEC = 180^o
\Rightarrow\quad\;\;\angle AED = \angle DEC = 90^o\qquad\qquad\quad\text{(from above)}
Hence the diagonals bisect each other at right angles.
Further, since \triangle AED\cong\triangle CED, therefore by CPCTC,
\qquad\quad\angle ADE = \angle CDE
Hence \angle D is bisected by the diagonal BD. Similarly, all the interior angles are bisected by the diagonals.
Q.E.D.
Recommended:
Characteristics of parallelogram
Angle sum property of polygon
Area of rectangle
No comments:
Post a Comment