In the given figure, a
quadrilateral 𝐴𝐵𝐶𝐷 is drawn to circumscribe a circle, with centre 𝑂, such that
the sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴 touch the circle at the points 𝑃, 𝑄, 𝑅, and
𝑆 respectively. Prove that 𝐴𝐵 plus 𝐶𝐷 is equal
to 𝐵𝐶 plus 𝐷𝐴.
In order to answer this question,
we need to know that the sides of a polygon that circumscribe a circle are tangents
to that very circle. In this example, the line 𝐵𝐶 is a
tangent to the circle at the point 𝑄. And the line 𝐶𝐷 is also a tangent
to the circle at the point 𝑅. We know that two tangents that meet
at a point are of equal lengths. This means that the line segments
joining the points 𝐴 and 𝑃 and the points 𝐴 and 𝑆 are of equal lengths.
Similarly, the line segments that
join 𝐵 to 𝑃 and 𝐵 to 𝑄 are also of equal lengths. We can create two more equations of
this form. 𝐶𝑅 is equal to 𝐶𝑄 and 𝐷𝑅 is
equal to 𝐷𝑆. This next step is going to seem a
little bit strange. We’re going to add all of these
four equations. Adding the left-hand sides, and we
get 𝐴𝑃 plus 𝐵𝑃 plus 𝐶𝑅 plus 𝐷𝑅. And adding the right-hand sides, we
can see that this is equal to 𝐴𝑆 plus 𝐵𝑄 plus 𝐶𝑄 plus 𝐷𝑆.
Looking at our diagram though, we
can see that the line 𝐴𝐵 is made up of the line segments 𝐴𝑃 and 𝐵𝑃. So 𝐴𝑃 plus 𝐵𝑃 must be equal to
𝐴𝐵. Similarly, the line that joins 𝐶
to 𝐷 is made up of the two line segments 𝐶𝑅 and 𝐷𝑅. So that means 𝐶𝑅 plus 𝐷𝑅 is
equal to the line 𝐶𝐷. Let’s continue this pattern on the
right-hand side of the equation. The line 𝐵𝐶 is made up of the two
line segments joining 𝐵 and 𝑄 and 𝐶 and 𝑄. This means that 𝐵𝑄 plus 𝐶𝑄 is
equal to 𝐵𝐶.
Finally, we can see that the line
joining 𝐷 to 𝐴 is made up of the two line segments joining 𝐴 and 𝑆 and 𝐷 and
𝑆. So 𝐴𝑆 plus 𝐷𝑆 is equal
𝐷𝐴. And that’s it. We’ve done it! We have proven that 𝐴𝐵 plus 𝐶𝐷
is equal to 𝐵𝐶 plus 𝐷𝐴 in this scenario.