Cyclic Quadrilaterals

Let us take a circle and a quadrilateral $ABCD$ in it. In the quadrilateral, let us try to expand the angle at $A$ by shifting $B$ to $B'$ and $D$ to $D'$. What is the consequence at the angle $C$ which is angle $B' C D'$ now? More we expand the angle at $A$, the angle at $C$ shrinks more. Similarly expansion of the angle at $C$ results in contraction of angle at $A$, keeping the vertices of the quadrilateral on the given circle.

In this increase/decrease situations at $A / C$, will there be not the situation, when the angles at $A$ and $C$ are equal? What is that situation? Is not that an interesting situation?  Let us now restart from this special situation. Make a measured increase in the angle at $A$. What is the measure of decrease in angle at $C$? Repeat the process with a new measured increases at $A$, make your observations. What do you discover? Write down your discovery, in the format of a theorem.

Using the theorem, the angle at $A$ determines the angle at $C$. In a new activity proposed, let us not change the angle at $A$ and so $B$ and $D$ don't move from their positions. Choose another point $C'$ on the circle.  By our theorem the angle at $A$ determines the angle at $C'$ also. What is the relation between the angles at $C$ and $C'$?  If we choose another point $C''$ on the arc $BD$, what do you observe about the angles in the same segment of a circle? Write down your discovery.


We discover that angles in same segment of a circle are all equal. So angles at $C$, $C'$, $C''$, ... are all equal. Even at the point E, the point on the circle obtained by extending $BO$ , the angle remains the same.

However, $EO =  AO$.  So  $\angle OEA = \angle AEO$.

Thus angle at the centre, i.e., $\angle AOB$, which is the sum of $\angle OEA$ and $\angle AEO$ is actually twice the angle $\angle AEO$.  Does it say something about the relation between the angle formed at the centre by a segment and angles at any point on the circle made by the same segment?

No comments:

Post a Comment