Problem 27: Incenter and Circumcircle of a Triangle

Problem 27

Let I be the incenter of the a triangle ABC and S be its circumcircle. Let D the midpoint of the arc BC of S, such that A and D are on the opposite sides of BC. Prove that:

DB = DI = DC

Here’s a simulation that illustrates the problem.

Try dragging the three vertices. Are DB, DI, and DC always equal? How can we prove the same?

Solution

Here are some things that you need to know to be able to solve the problems.

Incenter of a Triangle

The point of concurrency of the internal angle bisectors of a triangle is known as its incenter. Here’s a simulation that shows the incenter. Tap the play button to start the animation.

You can try dragging the vertices of the triangle. Are the angle bisectors always concurrent?

Angles Subtended by Circular Arcs

Let P and Q be any two fixed points on the circumference of a circle, whose center is O. Then, if R be any point on the circumference, such that O and R on the same side of PQ, then,

∠PRQ = (1/2)∠POQ

That is, the angles subtended by the arc PQ at the circumference is half the angle subtended by it at the center. Here’s a simulation that demonstrates this property.

Try dragging the points P, Q, and R. Is ∠PRQ always half of ∠POQ?

Angles Subtended by Chords of a Circle

If two chords of a circle are equal, then they subtend equal angles at its center. Here’s a simulation that demonstrates this property.

Use the slider to change the length of CD. You can also drag the point D to move the chord around. When CD equals AB, does ∠COD also equal ∠AOB?

Angles in the Same Segment of a Circle

Let P and Q be any two points on the circumference of a circle. Let A, B, C, … be arbitary points on the circumference such that they’re on the same side of PQ. Then,

∠PAQ = ∠PBQ = ∠PCQ = …

That is, angles subtended by PQ in the same segment of the circle are equal. Here’s a simulation that demonstrates this problem.

Try dragging any of the points. Are the the angles PAQ, PBQ, PCQ, and PDQ always equal to each other?

Given the above tools, here’s how you can proceed.

  1. Given that D is the mid point of BC, what does that tell you about the chords BD and CD?
  2. What can you say about the measures of ∠BAD and ∠CAD?
  3. Using the previous two points, can you say that A, I, and D are collinear? Why or why not?
  4. To prove that ID = IB, you need to prove that ∠DBI = ∠DIB? How can you do that?
  5. To prove that ID = IC, you need to prove that ∠DCI = ∠DIC? How can you do that?

That’s all from my side. Good luck!

Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.

Problem Source: CutTheKnot

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