Place the circumcenter inside or outside the triangle and draw three segments to connect it with the vertices. Use the definition of a circumcenter to identify the relationship between these segments.
See solution.
Practice makes perfect
Let's consider a right triangle â–³ ABC. Also, let P be the circumcenter of â–³ ABC. Since we do not know where P should be, we will place it inside the triangle.
Now we can draw three segments connecting P and the vertices of the triangle. By definition, the circumcenter is equidistant from the vertices of the triangle. Therefore, we can mark those segments as congruent. We will also label the newly formed angles.
By the Triangle Angle-Sum Theorem, the sum of angle measures in every triangle is 180^(∘). We also know that ∠A is a right angle, so it measures 90^(∘). Using this information, we can find the sum of m∠B and m∠C.
By the Angle Addition Postulate, the following relations are true.
m∠B &= m∠3 + m∠4
m∠C &= m∠5 + m∠6
Let's add these two equations by sides. We will also use the fact that the sum of m∠B and m∠C is 90^(∘).
m∠B + m∠C &= m∠3 + m∠4 +m∠5 + m∠6
&⇓
90^(∘)&= m∠3 + m∠4 +m∠5 + m∠6
In the diagram, there are three isosceles triangles: â–³ ABP, â–³ CBP, and â–³ ACP.
Therefore, we can substitute m∠3 with m∠2, m∠5 with m∠4, and m∠6 with m∠1.
90^(∘)&= m∠3 + m∠4 + m∠5 + m∠6
& ⇓
90^(∘)&= m∠2 + m∠4 + m∠4 + m∠1
From the diagram, we can also see that ∠1 and ∠2 add up to ∠A, which is a right angle. Hence, ∠1 and ∠2 are complementary, which means that the sum of their measures is 90^(∘). Let's use this information to solve the equation and find the measure of ∠4.
90^(∘) = m∠2 + m∠4 + m∠4 + m∠1
â–¼
Substitute 90^(∘) for m∠1 + m∠2 and evaluate
Earlier we concluded that ∠5 is congruent to ∠4, so its measure is also 0^(∘). This implies that P must be collinear with B and C. In other words, P must lie on side BC.
This way we have proven that the circumcenter of a right triangle always lies on one of the triangle's sides.
Extra
Placing P Outside â–³ ABC
In this part, let's place the circumcenter outside the triangle. Using similar reasoning, we can obtain the same result. Let's start by connecting P with the vertices of the triangle. As mentioned before, since P is a circumcenter, it is equidistant from the vertices.
Therefore, â–³ ACP, â–³ BCP, and â–³ ABP are isosceles triangles. Applying the Isosceles Triangle Theorem, we obtain the relations between the base angles of these triangles.
Triangle
Angles
â–³ ACP
m∠2 = m∠3 + m∠4
â–³ BCP
m∠4 = m∠5
â–³ ABP
m∠1 = m∠5 + m∠6
Let's substitute m∠5 with m∠4 into the third equation and add the resulting equation to the first one.
m∠2 &= m∠3 + m∠4
^+ m∠1 &= m∠4 + m∠6
m∠1 + m∠2 &= m∠3 + m∠6 + 2m∠4
From the diagram, we can see that ∠1 and ∠2 add up to a right angle ∠A, so ∠1 and ∠2 are complementary angles. Also, as calculated before, in a right triangle the sum of two not-right angles is 90^(∘). Hence, in △ ACB, ∠3 and ∠6 are complementary. m∠1 + m∠2_(90^(∘)) &= m∠3 + m∠6_(90^(∘)) + 2m∠4
&⇓
90^(∘) &= 90^(∘) + 2m∠4
Let's solve this equation to find the measure of ∠4.
As before, we obtained that ∠4 and m∠5 both measure 0^(∘). This means that P is on BC. As we can see, in either case we get that the circumcenter of a right triangle lies on the hypotenuse of the triangle.
