A circle touches all four sides of
a quadrilateral 𝐴𝐵𝐶𝐷. Prove that 𝐴𝐵 plus 𝐶𝐷 equals
𝐵𝐶 plus 𝐷𝐴.
The first thing we should do is
draw a diagram. It’s easiest to draw the circle
first and then the quadrilateral around it with its sides touching the circle. In other words, the sides of the
quadrilateral are tangents to the circle. We draw a tangent to the circle and
another one and another one. And we close it off with a fourth
side. Now, we trim the sides to get our
Now, we just need to label the
vertices. Our quadrilateral is 𝐴𝐵𝐶𝐷. And so we need to label the
vertices in that order. Note that we couldn’t, for example,
swab 𝐷 and 𝐶 or would have the quadrilateral 𝐴𝐵𝐷𝐶 instead. The order of the vertices really
does matter. Okay, so our diagram is
complete. We have a circle touching all four
sides of our arbitrary quadrilateral 𝐴𝐵𝐶𝐷.
We’re required to prove that 𝐴𝐵
plus 𝐶𝐷 — that’s the sum of the lengths of these two sides — is equal to 𝐵𝐶 plus
𝐷𝐴 — that’s the sum of the lengths of the other two sides. This question involves the length
of the sides of the quadrilateral. And those sides are tangents to the
And when we have a question which
is about the lengths of tangents, the following theorem should come to mind: the
lengths of tangents drawn from an external point to a circle are equal. So if we have a circle and an
external point that’s any point which is outside the circle, then the two tangents
drawn from these points to the circle are equal in length.
Of course, the tangent lines
continue in both directions. But we’re only interested in the
line segment from the external points to the point of contact with the circle. How does this help us? Well, let’s label the point of
contact of the tangent with the circle.
We’ll let 𝑃 be the point where the
side 𝐴𝐵 touches the circle, 𝑄 be where 𝐵𝐶 touches the circle, 𝑅 is where 𝐶𝐷
touches the circle, and 𝑆 is where 𝐴𝐷 touches the circle. Now, we can apply our theorem,
starting with the external point 𝐴.
The theorem tells us that 𝐴𝑃 and
𝐴𝑆 have the same length. Similarly, with the external point
𝐵, we can see that 𝐵𝑃 and 𝐵𝑄 are equal in length. With the external point 𝐶, we see
that 𝐶𝑄 and 𝐶𝑅 have the same length. And finally, with the external
point 𝐷, we see that 𝐷𝑅 and 𝐷𝑆 have the same length.
Now, let’s give these lengths names
to make it easier to prove our statement. We’ll let 𝐴𝑃 have a length of 𝑎
and then 𝐴𝑆 has a length of 𝑎 as well. Similarly, with 𝐵𝑃 and 𝐵𝑄, we
can let little 𝑏 be their lengths. 𝑐 is length of 𝐶𝑄 and 𝐶𝑅 and
𝑑 is the length of 𝐷𝑅 and 𝐷𝑆.
We are now ready to prove that 𝐴𝐵
plus 𝐶𝐷 is equal to 𝐵𝐶 plus 𝐷𝐴. We can write the side lengths of
our quadrilateral in terms of lowercase 𝑎, 𝑏, 𝑐, and 𝑑. As 𝑃 lies on the line segment
𝐴𝐵, 𝐴𝐵 has a length of little 𝑎 plus little 𝑏. Similarly, we can see that 𝐵𝐶 is
𝑏 plus 𝑐, 𝐶𝐷 is 𝑐 plus 𝑑, and 𝐷𝐴 is 𝑑 plus 𝑎.
Remember that we need to show that
𝐴𝐵 plus 𝐶𝐷 equals 𝐵𝐶 plus 𝐷𝐴. Well, we can write 𝐴𝐵 and 𝐶𝐷 in
terms of little 𝑎, 𝑏, 𝑐, and 𝑑. And similarly, we can write 𝐵𝐶
and 𝐷𝐴 in terms of little 𝑎, 𝑏, 𝑐, and 𝑑. And we can see now that these two
are equal. The only difference is the order in
which 𝑎, 𝑏, 𝑐, and 𝑑 are added. Of course, this does not affect
We have, therefore, proved that
𝐴𝐵 plus 𝐶𝐷 equals 𝐵𝐶 plus 𝐷𝐴. The key to this proof was the
theorem that the lengths of tangents drawn from the external points to a circle are
equal. Once we saw that this theorem
applied, the proof was straightforward. We’re not required to prove this
theorem to answer this question. We are allowed simply to state and
apply this theorem.