Angles Subtended By Arcs
$\underline{Introduction:}$
When we say angle subtended by an arc, there can be two types of angles that an arc can subtend. One type is the angle obtained by joining the two end points of the arc with the center. This is called the angle subtended by the arc at the center. The other type is when you take any point on the circumference, outside the arc, and join this with the two end points of the arc. The diagram below shows how they are drawn. The blue part is the arc that is subtending the angle.
The arc, $\overset{\frown}{RS}$, shown in blue in all three figures above, is the arc which is subtending the angle.
In Figure 1, $\overset{\frown}{RS}$ is less than the semi-circumference and is called a $Minor\ Arc$. Minor arcs subtend angles less than $180^\circ$ with the center, as shown in Figure 1.
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In Figure 2, $\overset{\frown}{RS}$ is greater than the semi-circumference and is called a $Major\ Arc$. Major arcs subtend angles greater than $180^\circ$ with the center, as shown in Figure 2.
In Figure 3, $\overset{\frown}{RS}$ is equal to the semi-circumference, and is called the semi-circle. A semi-circles subtend angles equal to $180^\circ$ with the center, as shown in Figure 3.
The $\angle ROS$, marked by the double arc, is called the $Angle\ subtended\ by\ \overset{\frown}{RS}\ at\ the\ center$, where point $O$ is the center of the circle in each case.
Now if we take any point $P$ on the circumference, outside $\overset{\frown}{RS}$, and join the endpoints of the arc $R$ and $S$ with $P$, then $\angle{RPS}\ is\ the\ angle\ subtended\ by\ \overset{\frown}{RS}\ on\ the\ circumference$.
One very important property of these angles are:
$\unicode{x201C}The\ angle\ subtended\ by\ an\ arc\ at\ the\ center\ is\ twice\ the\ angle\ subtended\ by\ it\ at\ the\ circumference\unicode{x201D}$
We will see the formal proof of this below. Although we have shown three separate drawings, for major arc, minor arc and semi-circle for ease of understanding, only the first diagram should be enough for the proof.
$\underline{Theory:}$
In a circle with center $O$ and any arc $\overset{\frown}{RS}$, and any random point $P$ on the circumference outside $\overset{\frown}{RS}$:
$\angle ROS = 2 \times \angle RPS$
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$\underline{Construction:}$
We draw the line joining points $P$ and $O$, and extend the line $PO$ to any random point $Q$.
$\underline{Proof:}$
Consider $\triangle OPS$, $\angle QOS$ is an external angle. We know that the external angle is equal to the sum of the two opposite internal angles, in this case $\angle SPO$ and $\angle OSP$.
$\texttip{\therefore}{therefore} \angle SPO + \angle OPS = \angle QOS$
$\triangle OPS$ is an isosceles triangle since sides $OP$ and $OS$ are the radii of the same circle, they are equal.
$\texttip{\therefore}{therefore} \angle SPO = \angle OPS$
Therefore we can write:
$\angle SPO + \angle SPO = \angle QOS$.
$\texttip{\therefore}{therefore} 2 \times \angle SPO = \angle QOS$........$eqn\ (1)$
Similarly in $\triangle OPR$, $\angle QOR$ is an external angle, and also, $\angle OPR = \angle ORP$.
$\texttip{\therefore}{therefore} 2 \times \angle RPO = \angle QOR$........$eqn\ (2)$
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Adding equations 1 and 2, we get:
$2 \times \angle SPO + 2 \times \angle RPO = \angle QOS + \angle QOR$
$\texttip{\therefore}{therefore} 2 \times(\angle SPO + \angle RPO) = \angle QOS + \angle QOR$
From the figure, we can deduce that $\angle SPO + \angle RPO = \angle RPS$, and $\angle QOS + \angle QOR = \angle ROS$
So, replacing this values we can write:
$2 \times \angle RPS = \angle ROS$
This is what we were expected to show.
Additional Observations (Corollaries):
$\underline{Corollary\ 1:}$
Look at the diagram below:
Unlike our previous theory, this time we choose many random points $P_1$, $P_2$, $P_3$ on the circumference.
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From our theory, we know that for every point $P_1$, $P_2$, $P_3$:
$\angle{RP_1S} = \frac{1}{2} \times \angle{ROS}$, and
$\angle{RP_2S} = \frac{1}{2} \times \angle{ROS}$, and
$\angle{RP_3S} = \frac{1}{2} \times \angle{ROS}$
Hence, we can also write:
$\angle{RP_1S} = \angle{RP_2S} = \angle{RP_3S}$
Therefore, we can conclude that:
$For\ a\ given\ \overset{\frown}{RS},\ the\ angles\ subtended\ at\ all\ points$$of\ the\ remaining\ circumference\ will\ have\ the\ same\ value.$
$\underline{Corollary\ 2:}$
Take a look at Figure 3 above. $\overset{\frown}{RS}$ is a semi-circle, so the chord $RS$ is a diameter,
$\texttip{\therefore}{therefore}\angle ROS = 180^\circ$
We also have seen that
$\angle{RPS} = \frac{1}{2} \times \angle{ROS}$
$\texttip{\therefore}{therefore} \angle{RPS} = \frac{1}{2} \times 180^\circ = 90^\circ$
Therefore, we can conclude that:
$The\ angle\ subtended\ by\ the\ semicircle\ or\ the\ diameter$$at\ the\ circumference\ is\ equal\ to\ 90^\circ.$