Question

Points A, B, C and D lie on circle O, and \[\left( {\overline {AC} } \right) \bot \left( {\overline {BD} } \right)\] at O. Prove that quadrilateral ABCD is a square.

Answer 1

Hint: Use the property of cyclic quadrilateral i.e. “Opposite angles of a cyclic quadrilateral are supplementary.”
We will first do the construction as given in the problem,
To prove $\square ABCD$ as a Square we will first prove it as a rectangular.
As shown in figure, AC is a diameter.

Therefore, $\angle ABC$ is a Semicircular angle, which will always be $90^\circ $
$\therefore \angle ABC = 90^\circ $………………………………… (1)
As $\square ABCD$ is a cyclic quadrilateral hence its opposite angle should be
Supplementary as it’s a property of cyclic quadrilaterals.
$\therefore \angle ABC + \angle ADC = 180^\circ $
$\therefore \angle ADC = 180^\circ - \angle ABC$
$\therefore \angle ADC = 180^\circ - 90^\circ $ [From (1)]
$\therefore \angle ADC = 90^\circ $
$\therefore \angle ABC = \angle ADC = 90^\circ $……………………..………………… (2)
Similarly we can prove,
$\therefore \angle BAD = \angle BCD = 90^\circ $………………………..……………… (3)
From (2) and (3) we can write,
$\therefore \angle ABC = \angle ADC = \angle BAD = \angle BCD = 90^\circ $
Since all angles of $\square ABCD$ are $90^\circ $, therefore it’s a Rectangle………………………(4)
Assume the radius of the circle is r.
It is given that \[\left( {\overline {AC} } \right) \bot \left( {\overline {BD} } \right)\]
Therefore $\vartriangle BOC$ is a Right angled triangle.
Therefore by Pythagoras Theorem,
$\mathop {OB}\nolimits^2 + \mathop {OC}\nolimits^2 = \mathop {BC}\nolimits^2 $
But, \[OB = OC = r\] as both is radius of circle,
\[\mathop {\therefore r}\nolimits^2 + \mathop r\nolimits^2 = \mathop {BC}\nolimits^2 \]
\[\therefore \mathop {BC}\nolimits^2 = \mathop {2r}\nolimits^2 \]
\[\therefore BC = \sqrt {\mathop {2r}\nolimits^2 } \]………………………………………. (5)
Also, \[\vartriangle AOB\] is a Right angled triangle,
Therefore by Pythagoras Theorem,
$\mathop {OB}\nolimits^2 + \mathop {OA}\nolimits^2 = \mathop {AB}\nolimits^2 $
But, \[OB = OA = r\] as both is radius of circle,
\[\mathop {\therefore r}\nolimits^2 + \mathop r\nolimits^2 = \mathop {AB}\nolimits^2 \]
\[\therefore \mathop {AB}\nolimits^2 = \mathop {2r}\nolimits^2 \]
\[\therefore AB = \sqrt {\mathop {2r}\nolimits^2 } \]……………………………………. (6)
From (5) and (6),
\[\therefore AB = BC\]………………………………………. (7)
I.e. the adjacent sides of a quadrilateral are the same.
Therefore from equation (4) and (7) we can say that $\square ABCD$ is a Rectangle with equal adjacent sides.
And any rectangle whose adjacent sides are equal is a Square.
Therefore $\square ABCD$ is a Square, Hence proved.
Note:
1. The concept of semicircular angle is very much important to prove the problem statement.
2. Always remember that the square is nothing but a rectangle with adjacent sides equal therefore if you prove the quadrilateral as a rectangle then it will be very much easier to prove it as a square.