Prove that the tangents drawn from
an external point to a circle are equal in length.
Let’s sketch a diagram. We can draw a circle with center 𝑂
with two tangents drawn from an external point 𝑃. Let’s also call the points at which
these tangents meet the circle 𝐴 and 𝐵.
Adding in the line segments 𝑂𝐴
and 𝑂𝐵, we can see that we have constructed two radii for our circle. Since these are the radii of the
same circle, we can deduce that the lines 𝑂𝐴 and 𝑂𝐵 are of equal length. We also know that the radius and
tangents meet at 90 degrees. So angle 𝑂𝐴𝑃 and 𝑂𝐵𝑃 are both
Let’s add one more line into our
diagram. That’s the line segment joining 𝑂
to 𝑃. We have now created two triangles
𝑂𝐴𝑃 and 𝑂𝐵𝑃 that have a shared line 𝑂𝑃. We have already shown that the
sides 𝑂𝐴 and 𝑂𝐵 are of equal length and that angle 𝑂𝐴𝑃 is equal to angle
𝑂𝐵𝑃, which is equal to 90 degrees.
The triangles have a right angle
each, a hypotenuse of equal length, and one other side is also of equal length. This is a condition of congruency,
often called RHS, where R stands for right angle, H stands for hypotenuse, and S
stands for side.
We can, therefore, deduce that the
triangles are congruent; that is to say, their sides are all of equal length. Since these two triangles are
congruent, it follows that 𝐴𝑃 must be equal in length to 𝐵𝑃.
Remember we said that the lines
joining 𝐴 and 𝑃 and 𝐵 and 𝑃 were the tangents to the circle. So we have proven that the tangents
drawn from a point to the circle are of equal length.