Basic and Rich Geometry Concepts part 4, Proof of arc angle subtending concept

Proof and uses of rich concept of angle subtended by minor and major arcs in a Circle

Basic and rich geometry concepts part 4 proof arc angle subtending

Arc angle subtending concept

To start with we will be using the following figure to explain the concept. In the figure an arc AB, colored red, subtends an angle $\angle APB=\theta$ at any general point P on the complementary arc APB that is colored blue. This complementary arc is the major arc in this case. The same red colored portion of arc AB subtends an angle $\angle AOB$ at the centre.

Basic and rich geometry concepts part 4-1 proof arc angle subtending

With this background now we will state the well known relation between the two angles, $\angle APB$ and $\angle AOB$. For convenience we call this relation as the Arc angle subtending concept.

This important concept states,

Any arc in a circle will subtend an angle at the centre twice the angle it subtends at any point on its complementary arc.

By this definition, in the above figure, the minor or smaller arc red colored AB subtends an angle $\angle AOB=2\theta$ at the centre where the angle it subtends at any general point P on its complementary major arc APB is, $\angle APB=\theta$.

Instead of the terms major or minor arc we will use from now on, a single terminology of Complementary arc that forms the rest of the circle, reference being the arc of central interest. In our case, arcs red colored AB and blue colored APB are complementary to each other.

The figure above shows the first and generally encountered form used for explanation of this concept.

In our previous geometry basic and rich concepts part 3 circles we have mentioned this rich concept and later we have used it extensively for solving geometry problems of diverse difficulty levels. But we have not discussed how this relationship holds true. In other words we have not shown the proof of this concept.

In this session we will discuss the mechanism by which the relationship holds true in all configurations.

You may skip the reasons for the need shown below.


Need of discussing the proof in all forms

We feel the need of showing exhaustively how this happens because of three reasons,

  1. This property of subtended angles in a circle though is of basic nature, it is used extensively in solving geometry problems. Unless you know the mechanism of how this happens, you won’t have inherent belief on the concept and will tend to apply the concept mechanically like a formula memorized by heart.
  2. Not only is this concept used directly to solve problems, it is also used as a basis for other rich and important concepts in geometry. This makes the concept still more important.
  3. Lastly, we tend to deal with the simplest configuration as we have shown above to explain this important concept. Unless such an important concept is shown in all its variations and its mechanisms, while solving problems with unusual configurations where you need to use the mechanism, you might hesitate because of lack of total clarity. Exhaustive understanding of how the concept works in all its variations should create complete clarity of the concept along with its mechanisms in your mind thus increasing confidence and effectiveness of use of the concept.

First configuration

The above figure represents the first configuration. The specialty of this simplest configuration is, when you join the centre with the point P on the complementary arc, two nice looking isosceles triangles are formed on two sides of the radius joining the centre and point P aiding the logic proof greatly.

Mechanism

(or proof, this term 'proof' we don't like, that's why it is secondary here)

Let's assume in two isosceles triangles $\triangle APO$ and $\triangle BPO$ (isosceles because radii AO=BO=PO),

$\angle APO = \angle OAP = x$ and

$\angle BPO = \angle OBP =y$.

Here, $x + y = \theta = \angle APB$, angle held by arc AB at a point P on the complementary arc.

So in these two triangles,

$\angle AOP = 180^0 - 2x$, and

$\angle BOP = 180^0 - 2y$.

Adding the two,

$\angle AOP + \angle BOP= 360^0 - 2(x+y)=360^0 - 2\theta$.

But the relation between the three angles around the centre is,

$\angle AOB = 360^0 - (\angle AOP + \angle BOP) = 2\theta$, where $\angle AOB$ is the angle held by the arc AB at the centre.

Thus angle held at the centre by the arc AB is double the angle it holds at any point P on the complementary arc APB.

Two sub-concepts

There are two concepts here:

  1. the angle subtended by the arc $AB$ at the center is twice this angle subtended at the periphery at any point (on the complementary arc).
  2. the arc $AB$ subtends the same angle at any point on the rest of the periphery of the circle (on the complementary arc). If the first is true the second follows immediately.

Now if you rotate P along the periphery, say anti-clockwise (clock-wise rotation results will be mirror image and same), at a certain stage, the points P, centre O and the chord end B will fall into a straight line. Prior to that configuration, the configuration in the above figure and its corresponding proof will hold good at all points on the complementary arc.

This is our second configuration and here instead of the previous mechanism, a new mechanism will still make relation hold true.

Second configuration

This is an important configuration as an additional geometric concept that is very basic will hold true in this configuration. The figure is as shown below.

Basic and rich geometry concepts part 4-2 proof arc angle subtending

As B, O and P are collinear, the three points form a diameter BP.

Mechanism

External angle $\angle AOB = \angle OPA + \angle OAP = \theta + \theta = 2\theta$, where $\angle APO = \angle APB = \theta$ held by the arc AB on the complementary arc APB. result is same, only the mechanism is different.

Diameter subtends an angle of $90^0$ at the periphery

This is the additional result we were talking about.  What is the mechanism here? Again it is the same arc angle subtending concept at  work here. Only, here the arc is the major arc on the opposite side of the diameter and this holds an angle of $180^0$ at the centre so that the angle held by the same major arc on its complementary arc is half of $180^0$, that is, $90^0$. This is the mechanism behind the well known result.

If we now continue to rotate the point P anti-clockwise along the periphery of the circle we reach a third configuration where we need to use yet again a new mechanism to show how the concept works.

Third configuration

The third position of the point P is shown in the figure below.

Basic and rich geometry concepts part 4-3 proof arc angle subtending

In this case also two isosceles triangles $\triangle OAP$ and $\triangle OBP$ are formed with base angles, $\angle OPA = \angle OAP = y$ and $\angle OPB = \angle OBP = x$.

Furthermore we assume for brevity, angle subtended by the arc AB at a point P on the complementary arc is $\angle APB = a$ and angle subtended at the cente by the same arc, $\angle AOP = b$. Here point P is on the same side of the diameter extending radius BO as the arc AB is. In the first configuration, P was on the opposite side of the diameter and in second configuration, on the diameter itself. This difference in relative positions of the point P created three different configurations.

In the two triangles involved,

$\angle AOP = 180^0 - b - 2x = 180^0 - 2y$

Or, $b = 2(y - x) = 2a$.

In this configuration also the same relation holds good though the signs of some angles are different.

These three configurations then makes us confident that at all points P on the complementary arc of AB the arc angle subtending concept will hold good.

Now if we continue rotating point P ant-clockwise further it will cross A and will lie between the points A and B. This is the fourth configuration where the relationship between the angles will still hold true, but the definition of the angles will be different. This creates a complementary concept.

Fourth configuration

This fourth position of the point P is shown in the figure below.

Basic and rich geometry concepts part 4-4 proof arc angle subtending

Here also we have two isosceles triangles and the angle relationships are,

$\angle p = 180^0 - 2\angle x$, and

$\angle q = 180^0 - 2\angle y$.

Adding the two we get,

$\angle p + \angle q = 360^0 - 2(\angle x + \angle y)$

Or. $\angle AOP = 360^0 - 2\angle \alpha$,

Or, $360^0 - \angle AOP = 2\angle \alpha$

Or, $\angle \beta = 2\angle \alpha$,

Here $\angle \beta$ is the angle held by the major arc at the centre and $\angle \alpha$ is the angle held by the same arc at any point P on its complementary mintor arc APB.

Again a similar relationship between arc subtended angles holds, but in this case the $\angle \beta$ is the starting angle held at the centre by the arc of interest as the major arc.

As per our arc angle subtending concept, this angle is double the angle held by the same major arc on any point P of its complementary arc which happens to be the minor arc. We have used minor arc as the reference arc in our three previous configurations with the major arc as the complementary arc. In this fourth configuration, the roles are reversed and the major arc is now the reference arc and the minor arc is the complementary arc.

In spite of the role reversal though, the arc angle subtending concepts holds good.

Note: If we use the terms, major arc and minor arc, this fourth configuration needs a role reversal between major arc and minor arc in defining the concept and thus a second complementary definition. But if we use the concept of complementary arc, a single definition is sufficient for all purposes.

Use of the concept

We identify four different rich concepts that are based on the arc angle subtending concept,

  1. Angle subtended by an arc on any point of its complementary arc is same. It follows from the angle subtended concept because this angle equals a fixed angle twice its value at the centre. This is an important concept.
  2. A diameter subtends and angle of $90^0$ at any point on the periphery (on both sides).
  3. Rich concept of Cyclic quadrilaterals in which sum of opposite angle will always be $180^0$. This is explained in our previous discussion on geometry basic and rich concepts part 3 circles.
  4. Rich concept of Secant of a tangent. This again is a rich concept that uses angle subtending concept as a basic mechanism. This also has been explained in our previous discussion on geometry basic and rich concepts part 3 circles.

Exhaustive approach

By following the Principle of exhaustivity we have thus devised a method of rotating the point P along the periphery of the circle to form all possible unique configurations related to the arc angle subtending concept for understanding the mechanism behind it exhaustively and comprehensively.

At the end we will solve a selected problem that uses the third special configuration of the concept of arc angle subtending. Before going though the solution you should try to solve the problem on your own.

Problem

Two chords AB and CD of a circle with centre at O intersect at point P so that overlapping $\angle AOD=100^0$ and $\angle BOC=70^0$. The value measure of $\angle APC$ is then,

  1. $80^0$
  2. $95^0$
  3. $65^0$
  4. $15^0$

Solution

Problem analysis

The figure depicting the problem is shown below,

Basic and rich geometry concepts part 4 problem proof arc angle subtending

The desired angle to be found out is $\angle APC$. This being one of the four angles at the intersection point P of the two chords AB and CD, out of four angles if one angle is known rest can be found out (because four angles consists of two unique angles only).

The $\angle APD$ being in the region of the given angles, we feel this should give us the breakthrough.

Sure enough when we look closely we locate two instances of arc angle subtending concept,

$\angle AOC=2\angle ADC$, and

$\angle DOB=2\angle BAD$, whereas,

$\angle APD = 180^0 - (\angle ADC + \angle BAD)$, in $\triangle APD$

$\hspace{15mm}=180^0 - \frac{1}{2}(\angle AOC + \angle DOB)$

$\hspace{15mm}=180^0 - \frac{1}{2}(100^0 - 70^0)$

$\hspace{15mm}=180^0 - 15^0$

$\hspace{15mm}=165^0$.

Thus,

$\angle APC = 180^0 - \angle APD = 15^0$.

Answer: Option d: $15^0$.

Key concepts used: End state analysis -- two line intersection property -- arc angle subtending concept -- angle overlapping pattern identification.


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