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A Lesser-Known Theorem for Circles

by Economist GMAT on May 23rd, 2012
6 comments
Click here to learn more about The Economist GMAT Tutor and its products & services.
Posted in
  • Circles
  • Geometry
  • GMAT Math
  • Lines and Angles
  • Problem Solving

Virtually everyone has committed a few circle formulas to memory.

We all know the formula for calculating the area of a circle (Area = pi*r^2), and most of us also remember how to compute its circumference (Length = 2*pi*r). There is, however, much more to circles than these two equations. Other properties are equally useful on the GMAT, if not so widely known. One such property is the inscribed angle theorem, which describes a special relationship between angles and circles. Because this theorem features prominently on the GMAT, it is well worth reviewing.

If we start with a basic circle, we can create an angle by joining the center of the circle to two radii. The resulting angle is known as a central angle, since the vertex of the angle is the center of the circle itself (the other two points lie on the circumference). The diagram below features circle O, where its two radii, OA and OB, join together at the center to form central angle ∠AOB:

A Central Angle

Angle ∠AOB forms a central angle

Incidentally, every central angle also defines an arc. In the diagram above, central angle ∠AOB defines the arc AB. An arc can actually be measured in two separate ways: by length and by angle. Much like a line, an arc has a distance that it spans, and so its length can be measured. Much like a central angle, however, an arc can also be measured in degrees, in which case the arc simply has the same value (in degrees) as the central angle which defines it. Arc AB is simply x°.

An Inscribed Angle

Angle BCA is an inscribed angle

There are also other interesting angles that do not have a vertex at the center. One special type of circle-related angle has all of its three points lying on the circumference. If we connect these points, we end up with an inscribed angle, like ∠BCA to the right:

An inscribed angle essentially consists of two chords — line segments that lie on the circle — connected to a point that lies on the circumference.

The Inscribed Angle Theorem

The inscribed angle theorem

It turns out that there is an interesting proof that relates an inscribed angle with the measure of its arc. Known as the inscribed angle theorem, it states that the measure of the inscribed angle is simply half the measure of the arc it defines. This property is absolutely invaluable for helping us solve geometry problems on the GMAT:

Since arc AB measures 2x, an inscribed angle with that same arc will measure exactly half that (which is simply x).

The Inscribed Angle Theorem (2)

An inscribed angle is half its corresponding central angle

There is an alternative way to express the inscribed angle theorem. Since a central angle and the arc it defines are equivalent, the following diagram is also true:

According to the inscribed angle theorem, the inscribed angle (x) is exactly ½ the central angle (2x).


To see how we can use this helpful theorem on the GMAT, let’s look at the following problem to the right:

Problem #1

If BD=DC, what is the value of

If BD=DC, what is the value of ‘x‘ in the figure to the right?

We’d like to take advantage of the inscribed angle formula, but as you’ll notice, ∠BAD and ∠BCD don’t share the same arcs, so we can’t apply the formula to relate the two angles. Let’s fix that by adding some new line segments. We’ll connect points B and C to form the isosceles triangle BCD. The base angles of this triangle are (180°-30°)/2=75° each. Using the inscribed angle theorem, we then find the measure of the arcs defined by each inscribed angle.

Problem #1 Redrawn

Connect points BC to reveal new inscribed angles

Inscribed angle ∠CBD is 75°, so arc CD is 150°; similar logic helps us find the measure of arc BAD. We can also apply the inscribed angle formula for ∠BDC to find that arc BC is 60°.

At this point, we can calculate the measure of arc BCD = 60 + 150 = 210. Since inscribed angle BAD defines arc BCD, the inscribed angle must be half of 210, so that x = ½×(210°) = 105°.


Problem #2

In the regular octagon above, what is the value of central angle

Now that you’ve learned how to use the inscribed angle theorem, I’d like to leave you with an interesting challenge problem:

In the regular octagon to the right, what is the value of central angle x?

If you liked this article, let Economist GMAT know by clicking Like.

RELATED ARTICLES

6 comments

  • Shruti on May 23rd, 2012 at 12:05 am

    The central angle X = 135

    Approach :
    Each interior angle of the octagon is [(8-2)*180]/8 = 135 degrees

    Consider the angles in each of the two triangles(isosceles) = 135 + 45(the central angle formed by the triangle)
    Hence the the central angle formed by two triangles sum up to 45 + 45 = 90

    As the centre angles should add to be 360.

    360 - 90 = 270

    Now X = 270/2 = 135 degrees.

    Reply to this comment
  • sandeep on May 23rd, 2012 at 3:12 am

    Hi Shruti,

    How did u get 45 as the central angle measure?

    Pls explain!

    Thanks,

    Reply to this comment
    • Vinayak on May 23rd, 2012 at 4:38 am

      I think one other way and maybe an easier way to look at this problem is by taking into account the cyclic nature of the regular polygon. Since this is a regular octagon it canbe inscribed in the circle. we all know the measure of a circle is 360 degrees. Further you can see it forms 8 isocless triangle with two sides equal to radis of the circle. if so, then that 360 is comprised of 8 equal angles each equal to x (x=360/8=45) and so the correct answer is 3/8X 360 = 45X3= 135.

      Make sense ?

    • Atul Swami on May 23rd, 2012 at 9:30 am

      Hi Sandeep,

      Since this is a regular octagon. Each cord will inscribe a central angle of 45 degrees.
      360 degrees/8 sides = 45 degrees.

      And since 3 chords are inscribing central angle "x" of the imaginary circle, which inscribes) around the octagon.

      45 degrees * 3 chords = 135 degrees.

      Regards,
      Atul

  • Vinoth@GMAT Kolaveri on May 24th, 2012 at 7:29 pm

    i ve posted the solution here
    http://www.beatthegmat.com/challenge-problem-t112978.html#475212

    http://postimage.org/image/ujw8pzg8x/

    Reply to this comment
  • subhakam on February 26th, 2013 at 4:03 pm

    Hello Economist GMAT - i did not understand the solution - isn't angle X = <BAD formed from teh same arc /same chord BD??? Doesn't the same chord form the same angle anywhere across wherever the point lies on the circle? In other words isn't <BAD = <BCD?? I am confused - please help !!

    Reply to this comment

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