# How to prove that |z_1|*|z_2|=|z_1*z_2|, Complex Numbers

In this post I’ll be showing you how to prove that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

Firstly, let’s say that:

${ z }_{ 1 }=x+iy$

${ z }_{ 2 }=p+iq$

If this is the case, then according to the rules of complex numbers:

$\left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

$\left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

Secondly, let’s determine what ${ z }_{ 1 }{ z }_{ 2 }$ is…

${ z }_{ 1 }{ z }_{ 2 }\\ \\ =\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px+iqx+ipy-qy\\ \\ =\left( px-qy \right) +i\left( qx+py \right)$

As you can see, we get the result above – which is another complex number.

This means that:

$RHS\\ \\ =\left| { z }_{ 1 }{ z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| { \left| { z }_{ 2 } \right| }\\ \\ =LHS$

Therefore we’ve proven that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

You can watch a video related to this proof below…

# Useful trigonometric formulas for finding areas of circles

If you’re trying to find the area of a circle using integration methods, then these trigonometric formulas are going to be very useful:

First formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\cos ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }$

Second formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) }$

These formulas are to be used when you have to transform the expression:

${ y }=\sqrt { { r }^{ 2 }-{ x }^{ 2 } }$

You can either make:

$x=r\sin { \theta }$

Or…

$x=r\cos { \theta }$

The choice is yours. 🙂

# Deriving the formula for an ellipse

In this post, I’ll be demonstrating how one can derive the formula for an ellipse from absolute scratch.

To derive the formula for an ellipse, what we must first do is create a diagram like the one below.

** Click on the image above to see it in full size.

Now, the first thing we’ve got to acknowledge here is that:

${ D }_{ 1 }+{ D }_{ 2 }=2a$

What we’re basically saying is that D_1 + D_2 is equal to the length from -a to a in the diagram above.

This formula can be understood by watching the video below…

These photographs can also help the formula sink into your mind…

Ellipse Image 1:

Ellipse Image 2:

Now, look at the diagram at the top of this page once again…

What you will notice is that:

${ \left( c+x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 1 }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }^{ 2 }={ c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }=\sqrt { { c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 } } \\ \\ { \left( c-x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 2 }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }^{ 2 }={ c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }=\sqrt { { c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 } }$

If this is the case, we can say that:

** Click on the image of the workings to see it in full size.

Alright, so far so good… Now, it turns out – if you look at the diagram at the top of this page carefully, you will discover that:

${ b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\\ \\ \therefore \quad { c }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$

And this ultimately means that:

${ a }^{ 4 }+\left( { a }^{ 2 }-{ b }^{ 2 } \right) { x }^{ 2 }={ a }^{ 2 }\left( { a }^{ 2 }-{ b }^{ 2 } \right) +{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ \therefore \quad { a }^{ 4 }+{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ -{ b }^{ 2 }{ x }^{ 2 }=-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ { b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }\\ \\ \frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\frac { { a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \\ \\ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\ \\ { \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1$

The formula you see just above is the formula for an ellipse. You’ve derived it from scratch!!

# Finding the formulas for areas of triangles

In this post I’ll be demonstrating how one can derive the three formulas which can be used to find the areas of triangles.

These formulas are in fact:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

To begin with, let’s start by looking at the diagram below:

Now, if you look at the diagram carefully – you will notice that the area of the triangle is:

$A=\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 }$

This can be simplified into:

$\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { x\cdot CN+\left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { CN\left\{ x+\left( c-x \right) \right\} }{ 2 } \\ \\ =\frac { CN\cdot c }{ 2 }$

Because of SOH CAH TOA, what we can also say is that:

$\sin { \left( A \right) } =\frac { O }{ H } =\frac { CN }{ b } \\ \\ \therefore \quad b\cdot \sin { \left( A \right) } =CN\\ \\ \sin { \left( B \right) =\frac { O }{ H } } =\frac { CN }{ a } \\ \\ \therefore \quad a\cdot \sin { \left( B \right) } =CN$

Now because:

$A=\frac { CN\cdot c }{ 2 }$

This ultimately means that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } \\ \\ A=\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) } \\ \\ \therefore \quad A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) =\frac { 1 }{ 2 } ac } \cdot \sin { \left( B \right) }$

Alright, so far so good… Now we must put the icing on the cake and attach the final piece of the jigsaw puzzle to the formula above. In order to find the three equations which can be used to find the areas of triangles, we must now discover the expression for sin(C). We can discover its expression by first saying that:

$C=\left( 90-A \right) +\left( 90-B \right) \\ \\ =90-A+90-B\\ \\ =180-A-B\\ \\ =180-\left( A+B \right) \\ \\ \therefore \quad \sin { \left( C \right) } =\sin { \left( 180-\left( A+B \right) \right) }$

And if we use the trigonometric identity below:

$\sin { \left( \alpha -\beta \right) } =\sin { \left( \alpha \right) \cdot \cos { { \left( \beta \right) } } -\cos { \left( \alpha \right) \cdot \sin { \left( \beta \right) } } }$

We will reach the conclusion:

$\sin { \left( 180-\left( A+B \right) \right) } =\sin { \left( 180 \right) \cdot \cos { \left( A+B \right) -\cos { \left( 180 \right) \cdot \sin { \left( A+B \right) } } } }$

But because:

$\sin { \left( 180 \right) =0 } ,\quad \cos { \left( 180 \right) =-1 } \\ \\ \sin { \left( 180-\left( A+B \right) \right) =-\left( -1 \right) \cdot \sin { \left( A+B \right) } } \\ \\ \therefore \quad \sin { \left( C \right) =\sin { \left( A+B \right) } }$

Now, sin(A+B) as a trigonometric identity, is:

$\sin { \left( A+B \right) =\sin { \left( A \right) \cdot \cos { \left( B \right) +\cos { \left( A \right) \cdot \sin { \left( B \right) } } } } }$

And, thanks to SOH CAH TOA…

$\sin { \left( A+B \right) =\sin { \left( C \right) } } \\ \\ \sin { \left( A \right) =\frac { CN }{ b } } \\ \\ \cos { \left( B \right) =\frac { A }{ H } } =\frac { \left( c-x \right) }{ a } \\ \\ \cos { \left( A \right) =\frac { A }{ H } =\frac { x }{ b } } \\ \\ \sin { \left( B \right) =\frac { CN }{ a } }$

Which means that…

$\sin { \left( C \right) =\frac { CN }{ b } \cdot \frac { \left( c-x \right) }{ a } +\frac { x }{ b } \cdot \frac { CN }{ a } } \\ \\ =\frac { CN\left( c-x \right) }{ ab } +\frac { CN\cdot x }{ ab } \\ \\ =\frac { CN\left( c-x \right) +CN\cdot x }{ ab } \\ \\ =\frac { CN\left\{ \left( c-x \right) +x \right\} }{ ab } \\ \\ =\frac { CN\cdot c }{ ab } \\ \\ \therefore \quad ab\cdot \sin { \left( C \right) =CN\cdot c } \\ \\ \therefore \quad \frac { 1 }{ 2 } ab\cdot \sin { \left( C \right) =\frac { CN\cdot c }{ 2 } =A }$

As this is the case, we can conclude that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

# Further Pure Maths: Complex Number Proof (1)

In this post, I’ll be proving that: $\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

First of all, let’s say that:

${ z }_{ 1 }=x+iy$

Whereby, $\left\{ x\in R,\quad y\in R \right\}$.

And also that:

${ z }_{ 2 }=p+iq$

Whereby, $\left\{ p\in R,\quad q\in R \right\}$.

If this is the case, this means that:

${ z }_{ 1 }\cdot { z }_{ 2 }=\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px-qy+i\left( qx+py \right)$

Therefore:

$LHS\\ \\ =\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+\left\{ { q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } \right\} } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \\ \\ =RHS$

Hence we’ve proven that:

$\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

# How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Ready? Let me begin…

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } }{ 2 }$

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

$A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

# 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? 🙂

# How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically

In this post I’ll be demonstrating how one can prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically…

First of all, let me show you this diagram…

sin(A-B)=sin(A)cos(B)-cos(A)sin(B) proof

*If you click on the diagram, you will be able to see its full size version.

IMPORTANT FACTS ABOUT THE DIAGRAM

Now, to begin with, I will have to write about some of the properties related to the diagram…

Property 1:

Angle B + (A – B) = B + A – B = A

Therefore, angle POR = A.

Property 2:

Angle OPS = 90 degrees

Property 3:

Length OS = 1

Also note:

All angles within a triangle on a flat plane should add up to 180 degrees. If you understand this rule, you will be able to discover why the angles shown on the diagram are correct. Angles which are 90 degrees are shown on the diagram too.

PROVING THAT SIN(A-B)=SIN(A)COS(B)-COS(A)SIN(B)

Since I’ve noted down some of the important properties related to the diagram, I can now focus on demonstrating why the formula above is true. I will demonstrate why the formula above is true using mathematics and the SOH CAH TOA rule…

$\sin { \left( A-B \right) } =\frac { O }{ H } =\frac { ST }{ 1 } =ST$

But it turns out that…

$ST=PR-PQ$

Because:

$QR=ST$

Now, what is PR and what is PQ?

$\sin { \left( B \right) } =\frac { O }{ H } =\frac { PS }{ 1 } =PS\\ \\ \cos { \left( B \right) } =\frac { A }{ H } =\frac { OP }{ 1 } =OP\\ \\ \sin { \left( A \right) } =\frac { O }{ H } =\frac { PR }{ \cos { \left( B \right) } } \quad \\ \\ \therefore \quad \sin { \left( A \right) } \cos { \left( B \right) } =PR\\ \\ \cos { \left( A \right) } =\frac { A }{ H } =\frac { PQ }{ \sin { \left( B \right) } } \\ \\ \therefore \quad \cos { \left( A \right) } \sin { \left( B \right) } =PQ$

And finally, to sum it all up:

$ST=PR-PQ\\ \\ \therefore \quad \sin { \left( A-B \right) =\sin { \left( A \right) } \cos { \left( B \right) } -\cos { \left( A \right) } \sin { \left( B \right) } }$

Need a better explanation? Watch this video…

Related Videos:

Related posts:

Simple But Elegant Way To Prove That sin(A+B)=sinAcosB+cosAsinB (Edexcel Proof Simplified)

# Properties of C squared, Pythagorean Theorem

In this post, I’ll be writing about some peculiar properties of C squared in Pythagoras’ theorem.

Look at this diagram very carefully…

*What are the weird properties of C^2..? It turns out that A1=A2 and A3=A4. A2 + A4 = C^2.

It turns out out that area A1 is equal to area A2, and that area A3 is equal to area A4:

A1 = A2

A3 = A4

This can be proven because:

1. ${ A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$
2. ${ x }^{ 2 }+{ D }^{ 2 }={ B }^{ 2 }$
3. ${ \left( C-x \right) }^{ 2 }+{ D }^{ 2 }={ A }^{ 2 }$

Now, due to the above:

${ D }^{ 2 }={ B }^{ 2 }-{ x }^{ 2 }\\ \\ { D }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ \therefore \quad { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-\left\{ { C }^{ 2 }-2Cx+{ x }^{ 2 } \right\} \\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx-{ x }^{ 2 }\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-\left\{ { A }^{ 2 }+{ B }^{ 2 } \right\} +2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ A }^{ 2 }-{ B }^{ 2 }+2Cx\\ \\ { B }^{ 2 }=-{ B }^{ 2 }+2Cx\\ \\ 2{ B }^{ 2 }=2Cx\\ \\ \therefore \quad { B }^{ 2 }=Cx\\ \\$

But… B^2 is actually the area A1 and Cx is the area A2, which means that A1=A2.

Now, if B^2=Cx, this means that:

${ A }^{ 2 }+Cx={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }={ C }^{ 2 }-Cx\\ \\ { A }^{ 2 }=C\left( C-x \right) \\ \\$

However, A^2 is equal to the area A3, and C(C-x) is equal to the area A4 – which means that A3=A4. Hence, we’ve proven that:

A1=A2

A3=A4

Related:

2 ways to derive Pythagoras’ equation from scratch

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation