Prove that the lengths of tangents drawn from an external point to a circle are even.
So what I’ve done first is actually drawn a sketch. So I’ve got a circle. Then, I’ve got two tangents to the circle and they actually meet at a point 𝑃, which is external. So it’s outside of our circle. So now, what I’ve also done is actually added the centre of the circle and I’ve called this 𝑂.
So to solve this problem, what we’re gonna have to do is actually prove that the length of tangents drawn from an external point to a circle are even. And what this means with our diagram is we have to prove that 𝑃𝑄 is equal to 𝑃𝑅. So to help us do that, I’ve drawn one more line, which is 𝑂𝑃. So now, what we have is actually two right-angled triangles and that actually we’ll explain what that means in a bit when we look at why they’re right angled. But we’re at this stage now.
Okay, so now, let’s go on and actually start to prove 𝑃𝑄 is equal to 𝑃𝑅. We’re gonna start by looking what’s shared between our two triangles. So okay, so first of all, we’ve got 𝑂𝑃 is equal to 𝑂𝑃. This is quite straightforward because this is actually common to both of our triangles. It is very important to actually say at each stage why you’re actually doing something. So here we’ve said that 𝑂𝑃 is equal to 𝑂𝑃 because it’s common to both triangles. So common to both triangles is our reasoning.
What we’ve also got is that 𝑂𝑄 is equal to 𝑂𝑅. And this is because they’re both the radii of our circle. And I’ve shown that because actually what I’ve put is a small 𝑟 to denote that. So we’ve got a radius at both point. Okay, so we’re doing well. We’ve actually shown that two things are common to our two triangles.
Then, the next thing we know is that both angle 𝑂𝑄𝑃 and 𝑂𝑅𝑃 are actually the same. So they’re equal to each other and that’s because they’re both 90 degrees. Now I mentioned this earlier. And the reason we know they’re both 90 degrees is that they are angles between a tangent and a radius. So when you got an angle between the tangent and the radius, then this is 90 degrees.
So therefore, we can say that actually triangle 𝑂𝑃𝑄 and triangle
𝑂𝑄𝑅 [𝑂𝑃𝑅] are gonna be congruent using the information that we found. And that’s because of RHS. And what RHS stands for is actually Right angle, Hypotenuse, Side. So both of our triangles have a right angle. So that’s correct. That’s the first part dealt with.
They also have a hypotenuse which is the same. And we know that because the hypotenuse is opposite the right angle. And in both of these cases, that’s 𝑂𝑃. And we’ve already said that 𝑂𝑃 is equal to 𝑂𝑃 because it’s common to both triangles. Okay, so that’s two points.
Let’s move on to the last bit, the side. Have they got a side the same? Well, yes, we know they’ve got a side the same because they both got one side which is the radius of our circle. So therefore, we can say that definitely triangle 𝑂𝑃𝑄 and triangle
𝑂𝑄𝑅 [𝑂𝑃𝑅] are congruent because of RHS. But how does this help us? Well, it helps us because what we’re trying to do is actually prove that 𝑃𝑄 is equal to 𝑃𝑅.
And now, we can actually say that 𝑃𝑄 is equal to 𝑃𝑅. And we’ve actually proved that because of CPCT. And it stands for Corresponding Parts of Congruent Triangles. And we know that first of all they’re congruent triangles cause we’ve already proved that. And then we know that they’re corresponding parts and that’s because of our congruent triangles.
We had 𝑅, which was the radius. So we had 𝑂𝑅 and 𝑂𝑄. So therefore, they’re gonna be already chosen and corresponding. Then, we also had 𝑂𝑃, which was shared. So that’s corresponding. So that leaves us with our final ones, which are 𝑃𝑄 and 𝑃𝑅. So therefore, these are the corresponding parts of our congruent triangles.
So there, we have it. We’ve actually proved that the length of tangents drawn from an external point to a circle are even because we’ve proved that 𝑃𝑄 is equal to 𝑃𝑅.