# Arcs and Subtended Angles

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An arc of a circle is any part of the circumference. The angle subtended by an arc at any point is the angle formed between the two line segments joining that point to the end-points of the arc. In the following figure, an arc of the circle shown subtends an angle $$\alpha$$ at a point on the circumference, and an angle $$\beta$$ at the center O:

We are interested in the relation between the angle which a given arc subtends anywhere on the circumference and the angle which that arc subtends at the circle’s center. The relation between them is simple, and extremely important.

Theorem: The angle subtended by an arc of a circle at its center is twice of the angle it subtends anywhere on the circle’s circumference.

The proof of this theorem is quite simple, and uses the exterior angle theorem – an exterior angle of a triangle is equal to the sum of the opposite interior angles. If the two opposite interior angles happen to be equal, then the exterior angle will be twice of any of the opposite interior angles.

Proof: Consider the following figure, in which an arc (or segment) AB subtends $$\angle {\rm{AOB}}$$ at the center O, and $$\angle {\rm{ACB}}$$ at a point C on the circumference.

We have to prove that$$\angle {\rm{AOB }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACB}}$$. Draw the line through O and C, and let it intersect the circle again at D, as shown:

We make the following observations:

1. In$$\Delta {\rm{OAC}},\angle {\rm{OAC }} = \angle {\rm{OCA}}$$ , because OA = OC

2. In$$\Delta {\rm{OBC}},\angle {\rm{OBC }} = \angle {\rm{OCB}}$$, because OB = OC

Thus,

$$\angle {\rm{AOD }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACO}},\angle {\rm{DOB }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{OCB}}$$

è$$\angle {\rm{AOD }} + \angle {\rm{DOB }} = {\rm{ }}2{\rm{ }} \times {\rm{ }}(\angle {\rm{ACO }} + \angle {\rm{OCB}})$$

è$$\angle {\rm{AOB }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACB}}$$

What if arc AB is such that it subtends a reflex angle at the center? Well, our proof doesn’t change, as the following figure shows:

You can now apply exactly the same steps as we did earlier, and arrive at the same result:$$\angle {\rm{AOB }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACB}}$$ .

Another configuration would be when O does not lie within$$\angle {\rm{ACB}}$$ . Will the result still hold, as in the following figure?

Yes, it will. Join C to O again and extend it to D on the circumference of the circle:

We note that (using the exterior angle theorem again):

1. $$\angle {\rm{BOD }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{BCD}}$$

2. $$\angle {\rm{AOD }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACD}}$$

To avoid cluttering, these angles have not been highlighted in the figure, but make sure you observe them carefully. Subtracting the two, we have:

$\angle BOD\;–\;\angle AOD\;\;=\;2\;\times\;(\angle BCD\;–\;\angle ACD)$

è $$\angle {\rm{AOB }} = {\rm{ }}2{\rm{ }} \times \angle {\rm{ACB}}$$

This theorem leads to an interesting corollary, which is discussed next.

Corollary: The angle in a semi-circle is a right angle.

Consider the figure below, where AB is a diameter of the circle. We need to prove that$$\angle {\rm{ACB }} = {\rm{ }}{90^0}$$ .

Proof: Since$$\angle {\rm{AOB }} = {\rm{ }}{180^0}$$,

$$\angle ACB{\rm{ }} = {\rm{ }}\raise.5ex\hbox{\scriptstyle 1}\kern-.1em/\kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\rm{ }} \times \angle AOB{\rm{ }} = {\rm{ }}{90^0}$$

This theorem also leads to the following result.

Theorem: Angles in the same segment of a circle are equal. In other words, an arc will subtend equal angles anywhere on the circumference.

Consider the following figure, which shows an arc AB subtending angles ACB and ADB at two arbitrary points C and D on the circumference. O is the center of the circle:

We need to prove that$$\angle {\rm{ACB }} = \angle {\rm{ADB}}$$.

Proof: Using our previous theorem, we have:

$\angle ACB\;=\;½\;\times\;\angle AOB,\;\;\angle ADB\;=\;½\;\times\;\angle AOB$

Thus,$$\angle {\rm{ACB }} = \angle {\rm{ADB}}$$.

Example 1: Find the value of x (in degrees) in each of the following figures (wherever indicated, O is the center of the circle).

I.

Solution: We note that that $$\angle ACB\;=\;\angle ADB\;=\;40^ 0 . Using\;\; the\;\; angle\;\; sum\;\; property \;\; in \;\; \triangle ACB$$, we have:

$\begin{array}{l}\angle BAC+\angle ACB+X^0=180^0\\\Rightarrow X^0=110^0\end{array}$

II.

Solution: $$\angle {\rm{BAD}}$$ is an angle in a semi-circle, and is therefore a right angle. Using the angle sum property in$$\Delta {\rm{ABD}}$$, we have:

$$\begin{array}{l}\angle ADB=180^0-(30^0+90^0)\\\;\;\;\;\;\;\;\;\;\;\;\;\;=60^0\end{array}$$

Thus,$$\angle ACB=\angle ADB=60^0$$.

III.

Solution: We have:

$$\begin{array}{l}\angle AOB=2\angle ACB\\\;\;\;\;\;\;\;\;\;\;\;\;\;=2(35^0+30^0)\\\;\;\;\;\;\;\;\;\;\;\;\;\;=130^0\end{array}$$

IV.

Solution: We note that

$$\begin{array}{l}\angle AOC=360^0-(150^0+120^0)\\\;\;\;\;\;\;\;\;\;\;\;\;\;=90^0\end{array}$$

Thus, $$\angle ABC=\frac12\angle AOC=45^0$$.

V.

Solution: We have:

$$\begin{array}{l}\angle BOC=180^0-40^0=140^0\\\Rightarrow X^0=\frac12\angle BOC=70^0\end{array}$$

VI.

Solution: The reflex angle AOB is equal to 3600 – 1500 or 2100. Thus, $$X^0=\frac12\times210^0=105^0$$

Example 2: In the following figure, find $$\angle {\rm{x}}$$ in terms of $$\angle {\rm{y}}$$ and $$\angle {\rm{z}}$$:

O is the center of the circle.

Solution: We label some angles as shown in the figure below:

Using the exterior angle theorem, we have:

$\begin{array}{l}\left\{ \begin{array}{l}\angle y = \angle 1 + \angle 2\\\angle 1 + \angle x = \angle 4\end{array} \right.\\ \Rightarrow \;\;\;\angle y + \angle 1 + \angle x = \angle 1 + \angle 2 + \angle 4\\ \Rightarrow \;\;\;\angle x + \angle y = \angle 2 + \angle 4\end{array}$ Now, since $$\angle {\rm{z }} = 2\angle 2{\rm{ }} = {\rm{ }}2\angle 4$$, we have$$\angle 2 = \angle 4{\rm{ }} = \angle {\rm{z}}$$ , and so:

$$\begin{array}{l}\angle x + \angle y = \angle z\\ \Rightarrow \;\;\;\angle x = \angle z - \angle y\end{array}$$

Example 3: Two circles intersect at P and Q. Through P, diameters are drawn for both the circles. The other end-points of these diameters are R and S. Show that R, Q and S are collinear.

Solution: Consider the following figure:

Since $$\angle RQP=\angle SQP=90^0$$  (angles in semi-circles), we have:

$$\begin{array}{l}\angle RQS=\angle RQP=\angle SQP\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;=180^0\end{array}$$

Thus, RQS is a straight line, which means that R, Q and S collinear.

Example 4: $$\Delta {\rm{ABC}}$$ is isosceles, with vertex A. Show that the circle on AB is diameter bisects BC.

Solution: Observe the following figure. The circle cuts BC at D:

Since $$\angle ADB=90^0$$  (angles in a semi-circle), AD is the altitude for the isosceles$$\Delta {\rm{ABC}}$$ , which means that AD also bisects BC, that is, D is the mid-point of BC.

Example 5: From the end-points of a diameter AB of a circle with center O, perpendiculars AP and BQ are dropped on an external line:

Show that AP = CQ.

Solution: We note that, $$\angle ACB=90^0$$ and thus ACQP is a rectangle, which means that AP = CQ.