বৃত্তস্থ কোণ উপপাদ্য

বৃত্ত সম্পর্কিত অনেক ধরণের উপপাদ্য আছে। তাদের মধ্যে অনেক cute একটি উপপাদ্য হলো বৃত্তস্থ কোণ উপপাদ্য।

উপরের চিত্রে আমরা একটি ত্রিভুজ ABC এবং তার পরিকেন্দ্র O। তাহলে, আমাদের উপপাদ্য বলে, $\angle BOC=2\angle BAC$ হবে। এখন আমরা এই উপপাদ্য প্রমাণ করব ।

প্রমাণ : এখানে, $\angle OAB=\angle OBA$ কারণ $OAB$ ত্রিভুজে $OA=OB$ এবং
$\angle OAC=\angle OCA$ কারণ $OAC$ ত্রিভুজে $OA=OC$ ।

এখন, $\angle BOD=\angle OAB+\angle OBA=\angle OAB+\angle OAB=2\angle OAB$......(i) এবং $\angle COD=\angle OAC+\angle OCA=\angle OAC+\angle OAC=2\angle OAC$......(ii)

(i) এবং (ii) থেকে আমরা পাই,

$\angle BOC=\angle BOD+\angle COD=2\angle OAB+2\angle OAC= 2(\angle OAB+\angle OAC)=2\angle BAC$

                                                 (প্রমাণিত)

এখন, উপপাদ্যটি যেহেতু আমরা শিখে ফেলেছি, আমাদের এখন এটি ব্যবহার করার সময়ে খেয়াল রাখতে হবে কোন কোণের দিগুন কোন কোণ হবে। তোমরা এখন নিচের প্রশ্নের উত্তর দিতে চেষ্টা করো

এখানে, প্রশ্নের উত্তর ২ নম্বরটি কারণ, এই উপপাদ্যটিতে, বৃত্তস্থ কোণ ($\angle BOC$) এবং কেন্দ্রস্থ কোণ ($\angle BAC$), উভয়ই একই চাপের ওপর অবস্থিত।

এই হচ্ছে আমাদের বৃত্তস্থ কোণ উপপাদ্য। 😁😁😁

Inscribed angles appear frequently in circle geometry:

(Here $\angle{ABC}$, with vertex $B$, is an inscribed angle of the green circle.)

Let's think about a natural question:

What's the angle measure of $\angle{ABC}$?

Of course, it will depend on the relative locations of $A,B,C$ on the green circle. A more precise question might be the following:

How, if at all, does the angle measure of $\angle{ABC}$ relate to other quantities in our diagram above?

If you're up for it, take a moment to think about this. The following "more complete diagram" might help. (Side note: to get to the heart of the matter (pun not intended), it's usually a good idea to draw the center of a circle---here, the point $O$. The basic reason is that central angles (like $\angle{AOB}$, with vertex at the center $O$) are the only angles we really know anything about yet---after all, we defined their angle size as arc length! If we then draw the radii $OA$, $OB$, $OC$, we get lots of symmetry and isosceles triangles. :)

Try playing around with this interactive Geogebra applet, too!

It turns out that if we fix points $A$ and $C$, the answer doesn't depend on $B$ (as long as $B$ stays on the same side of line $AC$---we'll look at what happens when $B$ "switches sides" soon)! In fact, $\angle{ABC} = \frac12 \angle{AOB} = \frac12 \widehat{AC}$. (Here arc $\widehat{AC}$ is interpreted as the arc from $A$ to $C$ that doesn't contain $B$. It's sometimes known as the arc intercepted by angle $ABC$.)

This is a very special property of the circle---for a "random curve/shape", we might expect the answer to depend on $B$ (i.e. change as $B$ moves around).

There are many nice ways to prove this result. If you're feeling adventurous, then in the spirit of Lockhart's Lament (p. 20-22), see if you can find a proof yourself! Otherwise see the follow-up post here.

Let's work through a couple proofs (hidden below the fold) of the result developed in the post here.

We start with the diagram from before. We wish to show $\angle{ABC}$ has half the measure/length of its intercepted arc $AC$ (the one not containing $B$), or $\frac12 \angle{AOC}$ (where we recall by definition that $\angle{AOC} = \widehat{AC}$).

(We've marked the radii $OA$, $OB$, $OC$ as segments of the same length. This gives us isoceles triangles $AOB$, $BOC$, $COA$ with vertices at $O$, so by the isosceles triangle theorem, we get $3$ pairs of equal angles, marked in red, yellow, and cyan. Recall that the reason we drew in the center and radii was to "get lots of symmetry and isosceles triangles".)

If you'd like a nice "angle chasing" exercise, try to show that $\angle{AOC} = 2\angle{ABC}$ yourself, before looking below.

Proof 1: direct but less enlightening

We know $\angle{ABC} = \text{blue} = \text{yellow} + \text{red}$. (@)

On the other hand, we can get a handle on $\angle{AOC}$ using the fact that the interior angles in a triangle sum to $\pi$ (or $180^\circ$ if you prefer). In triangle $AOC$ this gives $\pi = \angle{AOC} + 2\text{cyan}$. So $\angle{AOC} = \pi - 2\text{cyan}$. (#)

To relate the previous two equations, we use the same fact in triangle $ABC$ to get $\pi = 2\text{yellow} + 2\text{red} + 2\text{cyan}$. (*)

To finish, we combine these three equations together. Plugging (*) in (#) gives $\angle{AOC} = \pi - 2\text{cyan} = 2\text{yellow}+2\text{red}$. But by (@), this latter expression is just twice $\text{yellow}+\text{red} = \angle{ABC}$. In other words, $$\angle{AOC} = 2\angle{ABC},$$ which is what we wanted to prove.

Caveat to Proof 1

Here $B,O$ are on the same side of line $AC$. Does the proof above still work if $B,O$ are on opposite sides? If not, how does it need to be modified?

Proof 2: harder to come up with, but more enlightening

Here's a proof following the strategy that the most "fundamental" angles (or the "ones we know best") are the central angles: those with vertex at the center $O$ of the green circle.

The diagram looks complicated, but if we focus on the blue lines and angles, the proof almost goes without words. The idea is by shrinking everything by a factor of $2$ about $X$ (the reflection of $B$ over $O$, which also lies on the green circle), we've constructed a (blue) basically central angle $\angle{POR}$ visibly equal to the original $\angle{ABC}$ (since dilations preserve angles).

Now everything is centered at $O$, and we just need to show $\angle{AOC}$ is twice $\angle{POR}$. But ray $\overrightarrow{OP}$ cuts arc $\widehat{XA}$ in half (look at the red angles: we're using the congruent angles given by parallel lines $OP$ and $BA$). Similarly, ray $\overrightarrow{OR}$ cuts arc $\widehat{XC}$ in half (look at the yellow angles).

Thus gluing these two "half-arcs" together at $X$, we see that rays $\overrightarrow{OP}$ and $\overrightarrow{OR}$ together "halve" arc $\widehat{AC}$, so $\angle{POR}$ is half of $\frac12 \angle{AOC}$ (as usual, converting between arc length and central angle measure), so we're done.

The goal of the second proof was to explain the factor of $2$: why $\angle{ABC}$ is precisely half of $\angle{AOC}$ (or arc $\widehat{AC}$).

Caveat to Proof 2

Here $A,C$ are on opposite sides of line $BO$. Does the proof above still work if $A,C$ are on the same side? If not, how does it need to be modified?

(Hint: Proof 2 is basically the same as the proof on Wikipedia, just phrased differently. The Wikipedia article addresses both cases in full detail. How can we do the same here?)

There are some useful formulas to calculate angles in circles, formed by radii, chords, secants, and tangents.

1. Central Angle ========

A central angle is formed by two intersecting radii in a circle. Its vertex must be at the center of the circle.

In this diagram, x is the central angle. In a circle, or congruent circles, congruent central angles have congruent arcs.

Central Angle = Intercepted Arc

Thus,

Angle Measure of Arc AB = Central Angle x (Angle AOB)

1. Inscribed Angle =====

An inscribed angle is an angle with its vertex on the circumference of the circle. It is formed by two intersecting chords.

Inscribed Angle = 1/2 Intercepted Arc

Thus,

1/2*Angle Measure of Arc AC = Angle ABC

1. Tangent Chord Angle =====

An angle formed by an intersecting tangent and chord; the angle has its vertex on the circumference of the circle.

Tangent Chord Angle = 1/2 Intercepted Arc

Angle ABC= 1/2 Measure of Arc AB

1. Angle Formed Inside of a Circle By Two Intersecting Chords ======

When two chords intersect inside a circle, they created two sets of vertical angles (vertical angles are always equal). In the diagram, angle x is on of the vertical angles.

Angle Formed Inside Circle by Intersecting Chords = 1/2 Sum of Intercepted Arcs

Measure of Angle BED = 1/2 (Angle Measure of Arc AC + Angle Measure of Arc BD)

1. Angle Formed By the Intersection of Two Tangents, Two Secants, or a Tangent and a Secant: ========

The formula for ALL THREE of these cases are the same:

Angle Formed Outside Circle = |1/2 Difference of Intercepted Arcs|

a) Two Tangents

Angle ABC is formed by two tangents intersecting outside of the circle at point B. The intercepted arcs are minor arc AC and major arc AC.

b) Two Secants

Angle ACE is formed by two secants intersecting at point C. The intercepted arcs in this diagram are the minor arcs BD and AE.

c) A Secant and a Tangent

Angle ABD is formed by the tangent and secant intersecting at point B. The intercepted arcs in the diagram are the minor arcs AC and AD.

References: "Formulas for Working with Angles in Circles." Geometry Lesson Page. Regentsprep.org, 1988. Web. 6 Sept. 2014.

Angle - Arc - Relationship

Chords $AD$ and $AE$ intersect at point $A$ on the circle's circumference, forming $\angle{EAD}$. The #degrees in arc $DE$ is twice as much as the #degrees in $\angle{EAD}$.

Now you have learned about all parts of the circle except for tangents, so you can do this worksheet! (Yes, there are tangent questions, but if you know what you are doing, you can use process of elimination for them, and maybe even figure out what they are without reading the explanation!)

http://www.mathworksheets4kids.com/circles/parts-medium-large.png