# Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle… We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)? Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because: $\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn… As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that: $\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that: $x+y=180$

So, it turns out we now have two useful equations: $\alpha +\beta =180$ $x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves… • There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial: $\beta +y=180$ $\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof. $\alpha +\beta =180$ $x+y=180$ $\beta +y=180$ $\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that: $\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that: $\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y: # How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

If you look at this diagram carefully, what you will notice is:

• A circle exists and each point on this circle has the coordinate (x, y).
• The centre of the circle can be found at (a, b).
• The circle has a radius ‘r’.
• The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression: $\left( x-a \right)$ or $\left| x-a \right|$

The opposite lengths can be described using the expression: $\left( y-b \right)$ or $\left| y-b \right|$

Also, very interestingly:

• Each of the right angled triangles you can think of has a hypotenuse ‘r’.
• ${ \left( x-a \right) }^{ 2 }={ \left| x-a \right| }^{ 2 }$
• ${ \left( y-b \right) }^{ 2 }={ \left| y-b \right| }^{ 2 }$

When you combine all the information above, what you get is a neat formula which looks like this: ${ \left( x-a \right) }^{ 2 }+{ \left( y-b \right) }^{ 2 }={ r }^{ 2 }$

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? 🙂

# Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

Derive the formula of an ellipsehttps://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle: https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle: https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch: https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

If you love mathematical proofs, please feel free to join my ‘mathematics proofs’ Google community at: https://plus.google.com/communities/106007058741903558109

I’ve also got a new Facebook page related to stuff about the universe and also mathematics. You can add me as a friend by accessing this link: http://www.facebook.com/tiago.hands

# Integration Problem – Finding the area of quarter of a circle by integrating…

Integrate: $\int _{ 0 }^{ r }{ \sqrt { { r }^{ 2 }-{ x }^{ 2 } } } dx$

Say that: $x=r\sin { \theta }$

So… $x=r\sin { \theta } \\ \\ x=ru\\ \\ \frac { dx }{ du } =r\\ \\ u=\sin { \theta } \\ \\ \frac { du }{ d\theta } =\cos { \theta } \\ \\ \therefore \quad \frac { dx }{ d\theta } =r\cos { \theta } \\ \\ \therefore \quad dx=r\cos { \theta } d\theta$

Now… $When\quad x=r,\\ \\ r\sin { \theta =r } \\ \\ \sin { \theta } =1\\ \\ \therefore \quad \theta =\frac { \pi }{ 2 } \\ \\ As\quad \left\{ 0\le \theta \le \frac { \pi }{ 2 } \right\} \\ \\ When\quad x=0,\\ \\ r\sin { \theta } =0\\ \\ \sin { \theta } =0\\ \\ \therefore \quad \theta =0$

So you now have to integrate: $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }-{ { r }^{ 2 }\sin ^{ 2 }{ \theta } } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\cos ^{ 2 }{ \theta } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( { r }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } { \left( \cos ^{ 2 }{ \theta } \right) }^{ \frac { 1 }{ 2 } }\cdot r\cos { \theta } d\theta \\ \\ =\int _{ o }^{ \frac { \pi }{ 2 } }{ r\cos { \theta } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \cos ^{ 2 }{ \theta } d\theta$

But first, realise that: $\cos { \left( \theta +\theta \right) } =\cos { \theta } \cos { \theta } -\sin { \theta } \sin { \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -1+\cos ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1\\ \\ 2\cos ^{ 2 }{ \theta } =\cos { 2\theta } +1\\ \\ \cos ^{ 2 }{ \theta } =\frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 }$

So you now integrate: $=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \left( \frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 } \right) d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } { r }^{ 2 }\cos { 2\theta } +\frac { 1 }{ 2 } { r }^{ 2 }\quad d\theta$

And: $If\quad p=\frac { 1 }{ 4 } { r }^{ 2 }\sin { 2\theta } =\frac { 1 }{ 4 } { r }^{ 2 }u\\ \\ \frac { dp }{ du } =\frac { 1 }{ 4 } { r }^{ 2 }\\ \\ u=\sin { 2\theta } =\sin { q } \\ \\ \frac { du }{ dq } =\cos { q } =\cos { 2\theta } \\ \\ q=2\theta ,\quad \frac { dq }{ d\theta } =2\\ \\ \frac { du }{ d\theta } =2\cos { 2\theta } \\ \\ \therefore \quad \frac { dp }{ d\theta } =\frac { 1 }{ 4 } { r }^{ 2 }\cdot 2\cos { 2\theta } \\ \\ =\frac { 1 }{ 2 } { r }^{ 2 }\cos { 2\theta }$

Therefore: Now, to get the area of a circle, you multiply the final result by 4. $4\cdot \frac { 1 }{ 4 } \pi { r }^{ 2 }\\ \\ =\pi { r }^{ 2 }\\ \\ \therefore \quad A=\pi { r }^{ 2 }$

# Find The Formula For The Volume Of Cones

Find the volumes of cones formula using differentiation and integration. This playlist will show you how to come up with the volumes of cones formula from scratch. Knowledge about how to differentiate and integrate required.