The circumcircle and altitudes of a triangle

One of the most frequently used points in a triangle is its circumcenter - the center of the circle circumscribed about the triangle. We called this circle the circumcircle of the triangle. The circumcenter O is the intersection of the three perpendicular bisectors of the sides of the triangle. The sides of a given triangle ABC and its circumradius will be denoted as before by the letters a, b, c and R.

There are many interesting properties of the circumcircle connected with the altitudes of a triangle. On the Figure 1 we can see a triangle $\Delta ABC$, its circumcircle with center O and diameter AA₀ through A. We see also the altitude AD=h_a.

  

Figure 1: The circumcircle

Problem 1. $\Delta ABD \sim \Delta AA_0 C$. In particular $h_a=\frac{bc}{2R}$ and $\angle DAA_0=\vert\angle B - \angle C\vert$.

Proof. The fact that the two right triangles are similar follows by $\angle B=\angle AA_0 C$. So we have that

$$\frac{h_a}{c}=\frac{b}{AA_0}$$

and using that AA₀=2R we verify the first identity. For the second lets subtract from $\angle BAC$ the two equal angles

$$\angle A_0 AC=\angle BAD = 90^{\circ}-\angle B.$$

We are left with

$$\angle DAA_0 = \angle A - 2(90^{\circ}-\angle B)=\angle B-\angle C.$$

This expression for $\angle DAA_0=\angle DAO$ has been made with reference to the Figure 1 in which $\angle B>\angle C$. If instead we had taken $\angle B<\angle C$, the equal angles A₀AC and BAD would have overlapped, with the result that $\angle DAO=\angle C - \angle B$.

As we know the three altitudes AD, BE and CF have a common point - the orthocenter H of $\Delta ABC$. Here is a nice connection between H and the circumcircle.

Problem 2. The symmetric images of H with respect to the sides of $\Delta ABC$ lie on the circumcircle of $\Delta ABC$.

  

Figure 2: Problem 2

Proof. On the Figure 2 the three altitudes AD, BE and CF are extended to meet the circumcircle at D', E' and F'. Now $\angle DAB=\angle FCB$, both being complements of the angle B. Also $\angle BCD' = \angle BAD'$. So the right triangles CDH and CDD' are congruent by ASA and therefore HD=DD'. Similarly HE=EE' and HF=FF', which is sufficient for verifying the statement.

Similar to problem 2, but a little bit harder is the next problem.

Problem 3. The points P, Q and R on the sides BC, AB and AC of the isosceles triangle ABC (AB=AC) are such that PQ and PR are parallel to AC and AB respectively. Proof that the symmetric image of P with respect to the line QR lies on the circumcircle of $\Delta ABC$.

Proof. Denote with P' the symmetric image of P with respect to QR (see Figure 3). Since AQRP is parallelogram and because of the symmetry the distance between P and QR is the same as that between A and QR as well as between P' and QR. So $AP' \parallel QR$ and therefore $\angle APP' = 90^{\circ}$. To prove that P'.lies on the circumcircle we need to prove that $\angle B + \angle AP'C = 180^{\circ}$. It is sufficient to prove that $\angle B = \angle PP'C = 90^{\circ}$. But PR=RC (because AB=AC and $PR \parallel AB$) and PR=P'R (because of the symmetry) so it turns out that the point R is the circumcenter of $\Delta PP'Q$. So we have that:

  

Figure 3: Problem 3

$$\angle PP'C = \frac{1}{2} \angle PRC = \frac{1}{2} \angle BAC = 90^{\circ} - \angle B.$$