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What happens to the orthocenter as an angle a triangle goes from acute, to right, to obtuse?
See solution.
In the mentioned exercise we considered a triangle △ ABC. We also knew that in this triangle, the altitudes from B and C meet at point D, which makes D the orthocenter of △ ABC. We are also told that m∠ BAC > m∠ BDC. We found that the △ ABC is obtuse. Now, we take an acute, right, and obtuse triangle and check whether m∠ BAC or m∠ BDC is greater.
Let's draw an arbitrary acute triangle △ ABC, along with the altitudes from B and C. The point where they intersect is D.
This time, we draw an arbitrary right triangle △ ABC, along with the altitudes from B and C. The point where they intersect is D. This time, however, the point where the altitudes intersect is the vertex with the right angle.
In a right triangle, the orthocenter D is the same point as the vertex A. Therefore, m∠ BDC is equal to m∠ BAC, since they are the same angle — the right angle of the triangle.
Finally, we consider an arbitrary obtuse triangle, along with the altitudes from B and C.
This time, the orthocenter D is outside the triangle. This means that m∠ BAC is greater than m∠ BDC.
Now we apply the Triangle Angle Sum Theorem to each of these triangles. m∠ CBD + m∠ BCD + m∠ BDC = 180^(∘) m∠ CBA + m∠ BCA + m∠ BAC = 180^(∘) Note that, when △ ABC is acute and D is inside the triangle, m∠ CBD is smaller than m∠ CBA. Likewise, m∠ BCD is smaller than m∠ BCA. Therefore, for the sums of the interior angle measures to be equal, m∠ BDC must be greater than m∠ BAC.
However, when △ ABC is obtuse, D is outside the triangle and m∠ CBD is greater than m∠ CBA. Likewise, m∠ BCD is greater than m∠ BCA. Then, for the sums of the interior angle measures to be equal, m∠ BDC must me smaller than m∠ BAC.
