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
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