How to prove that the two angles below the apex of an isosceles triangle are equivalent

*You will need a pair of compasses, a ruler, pen and pencil to formulate this proof.

How would you go about proving that an isosceles triangle has two angles (below its apex) equal to one another?

Well, first of all – let’s start off by drawing a circle…

A circle.

Now… We can tell that the circle we’ve just drawn has a centre (point at the centre). Next, what we have to do is add a couple of points to the edge of this circle. Like this…

Circle with points

Let’s name all these points A, B and C…

Points a, b and c

Now, let’s connect these points together with a few lines – to create an isosceles triangle ABC…

Isosceles triangle

Ok… So far, so good… What you will need to do now is – place the needle of your compass on the point C and your pencil on the point B, like this…

Points B and C

Now spin your compass – and create an arc…

An arc...

Next, get the needle of your compass and place it on the point B and put your pencil on the point C…

Points B and C...

Draw another arc, like this…

2nd arc

Where the two arcs you’ve just drawn intersect, create a point… Call this point D…

Now, draw a line going through the points A and D. Call this line L. Line L will be perpendicular to the line BC…

Line going through points A and D...

Where the line L intersects the line BC, create a point E…

Point E

Now, it turns out, within the isosceles triangle ABC, we’ve created two right angles… This is because the line L is perpendicular to the line BC. Remember that the line L cuts the isosceles triangle down its centre. Let’s name these right angles big R…

Right angles within isosceles triangle...

If you look at the diagram above carefully, what you will notice is that the radius of the circle is equal in length to the line AB and also the line AC. Let’s name the lines AB and AC… We’ll call them r.

Circle contains radiuses...

Let’s also name the line BC… We’ll call it x. This means that the line BE is equal to half of x, and because of this, the line CE must also be equal to half of x…

Line x...

Finally (I know you must be tired of drawing), let’s call the angle ABC alpha and the angle ACB beta…

alpha and beta angles...

With our diagram complete, we can now prove that alpha and beta are equivalent to each other.

*You will need to know a bit of trigonometry to pass this point. SOH CAH TOA rules to be precise.

It turns out that:

\cos { \left( \alpha  \right)  } =\frac { A }{ H } =\frac { \frac { x }{ 2 }  }{ r } =\frac { x }{ 2r } 

And also:

\cos { \left( \beta  \right)  } =\frac { A }{ H } =\frac { \frac { x }{ 2 }  }{ r } =\frac { x }{ 2r } 

This means that:

\cos { \left( \alpha  \right)  } =\cos { \left( \beta  \right)  } \\ \\ \therefore \quad \alpha =\beta 

Hence, we’ve proven that an isosceles triangle has two angles (below its apex) equal to one another.

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