Problem 17
Let ABC be any triangle. Let H be the point of concurrency of its altitudes. Show that the image of H (i.e. the orthocenter) about any side of the triangle lies on its circumcircle.
The following simulation demonstrates the problem.
Try dragging the three vertices of the triangle. Do the images of H about each side always lie on the circumcircle? In other words, is HD = DP, HE = EQ, and HF = FR?
Solution
Here are a couple of things that you’ll need to know for this problem.
Congruent Triangles
If the corresponding sides and angles of two triangles are equal, then they are said to be congruent.
And, if we’re able to prove that two triangles are congruent, then their corresponding parts (sides and angles) are equal.
There are various ways to prove the congruence of two triangles, one of them being the Right Angle-Hypotenuse-Side or ASA rule.
That is, for two right triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = DE, then the two triangles would be congruent.
And, this would mean that ∠C = ∠F, BC = EF, and AC = DF (i.e., the corresponding parts are equal).
You can observe this in the following simulation.
Angles in the Same Segment of a Circle
The angles subtended by an arc in the same segment of a circle is constant. Here’s a simulation that demonstrates this property.
Try dragging one of the points A, B, C, or D and observe the measures of the angles PAQ, PBQ, PCQ, and PDQ. Are they always equal?
Now, to prove that P is the image of A about BC, we need to show that HD = DP and HP ⊥ BC. The latter is already true since we extended the altitude AD, meeting the circumcircle at P. To prove that HD = DP, you need to find two triangles that can be proven as congruent. Then, the corresponding parts of these triangles, HD and DP, would be equal. Good luck!
Were you able to solve the problem 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.
Previous Problem | Next Problem | Geometry Problems | Problems Home