### Video Transcript

Prove that the lengths of the
tangents drawn from an external point to a circle are equal.

For a question like this, it’s
always helpful to draw a diagram. So we’ll start by drawing a circle,
which we’ll call 𝐶, which has a centre of 𝑂. We can also mark the external point
to our circle, which is mentioned in the question. And we can label this point 𝑃.

We now need to draw tangents from
this external point to the circle. And we can remind ourselves the
definition of a tangent. The tangent to a circle is a line
which intersects the circle at exactly one point. In other words, the line just
touches the perimeter of the circle.

Looking at our diagram, we can draw
two lines which fit this description and these are the tangents which we’ll work
with. These tangents connect our circle
𝐶 with the point 𝑃. We can also label the points at
which these tangents touch the circle as 𝐴 and 𝐵. This allows us to refer to the
tangents as 𝐴𝑃 and 𝐵𝑃.

Now, the question is asking us to
prove that the lengths of these two tangents are equal. And in order to do so, we’ll need
to make a few more geometric constructions. To begin, we can connect the centre
of our circle with the two points 𝐴 and 𝐵. These two lines that we have just
drawn — 𝑂𝐴 and 𝑂𝐵 — are both radii of the circle since they connect the centre
𝑂 to the circumference via a straight line.

Given this information, we’re able
to say that the length 𝑂𝐴 is equal to the length 𝑂𝐵. And this is of course equal to the
radius of the circle 𝑟. Now that we’ve drawn these lines,
we can use our circle theorems to add more information to the diagram. You may be familiar with the fact
that when a tangent and a radius of the same circle are connected, they will always
be perpendicular.

Looking at our diagram, we can see
that this situation occurs twice at the points 𝐴 and 𝐵. We know that perpendicular lines
are at 90 degrees to each other. And we’re, therefore, able to mark
on a 90-degree angle between the lines 𝑂𝐴 and 𝐴𝑃 and between the lines 𝑂𝐵 and
𝐵𝑃. Now that we’ve marked angle 𝑂𝐴𝑃
and 𝑂𝐵𝑃 as 90 degrees, we can draw one final line on our diagram which connects
the centre of the circle 𝑂 with the external point 𝑃.

In doing this, we have created two
connected triangles: 𝑂𝐴𝑃 and 𝑂𝐵𝑃. We can see that both of these
triangles are right-angled triangles and they share a common side which is their
hypotenuse 𝑂𝑃. We now have that triangles 𝑂𝐴𝑃
and 𝑂𝐵𝑃 are both right-angled triangles. They share the hypotenuse and they
have one other side 𝑂𝐴 and 𝑂𝐵, respectively, which is equal in length.

This information is enough for us
to conclude that triangles 𝑂𝐴𝑃 and 𝑂𝐵𝑃 are congruent using RHS congruency. RHS congruency says that if the
hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and
one side of another right-angled triangle, then these two triangles are
congruent.

Now, we know that congruent
triangles have the same corresponding angles and the same corresponding side
lengths. This means that the third and final
side lengths of our triangles 𝐴𝑃 and 𝐵𝑃, respectively, must also be equal. Since the lines 𝐴𝑃 and 𝐵𝑃 are
the tangents, we have therefore answered the question and we have proved that the
tangents are equal.

As a quick aside, if you didn’t
know the RHS congruency rule, there is also another fairly straightforward way you
can solve this problem. We can do this using a familiar
tool, which is Pythagoras’s theorem. This tells us that for a
right-angled triangle, the square of the hypotenuse is equal to the sum of the
square of the other two sides.

In our case, for both of our
right-angled triangles, the length of the hypotenuse is 𝑂𝑃. For the triangle 𝑂𝐴𝑃, we can,
therefore, say that 𝑂𝑃 squared is equal to 𝑂𝐴 squared plus 𝐴𝑃 squared. By looking at triangle 𝑂𝐵𝑃, we
can also say that 𝑂𝑃 squared is equal to 𝑂𝐵 squared plus 𝐵𝑃 squared.

Let’s now observe this part of the
equation that has been formed. To move forward, we can recall that
earlier in our method, we proved that the length 𝑂𝐴 and the length 𝑂𝐵 were the
same since they’re both radii of the circle 𝐶. Since this is the case, we can also
see that by squaring both terms, 𝑂𝐴 squared is equal to 𝑂𝐵 squared.

Now that we know this, we look back
at our equation. And we can see that we can cancel
the 𝑂𝐴 squared on the left-hand side of our equation and the 𝑂𝐵 squared on the
right-hand side of our equation since both of these are equal. After doing so, the equation that
we’re left with is that 𝐴𝑃 squared is equal to 𝐵𝑃 squared. By taking the square root of both
sides of this equation, we once again arrived at the conclusion that 𝐴𝑃 is equal
to 𝐵𝑃.

We have now answered the
question. And we have proved using this
alternative method that the two tangents 𝐴𝑃 and 𝐵𝑃 are equal.