+61 1)Triangle ABC is an isosceles right triangle where angle A = 90 degrees. O is the circumcenter of Triangle ABC. What is angle AOB in degrees?
2)Let O and H be the circumcenter and orthocenter of acute triangle ABC, respectively. If angle AHC + angle AOC = 240 degrees, what is angle ABC in degrees?
+2446 I have been thinking about these two problems for some time, and I think I have finally cracked them! I'll now impart my knowledge to you!
1)
| \(\triangle ABC\text{ is an isosceles right triangle}\\ m\angle A=90^{\circ}\\ O\text{ is the circumcenter}\) | Given information |
| \(m\angle A+m\angle B+m\angle C=180^{\circ}\) | Triangle sum theorem (the sum of the interior angles of a triangle equals 180 degrees) |
| \(90^{\circ}+m\angle B+m\angle C=180^{\circ}\) | Substitution property of equality |
| \(m\angle B+m\angle C=90^\circ\) | Subtraction property of equality |
| \(m\angle B=m\angle C\) | Isosceles Triangle Theorem (The angles opposite congruent sides in an isosceles triangle are congruent) |
| \(m\angle B+m\angle B=90^\circ\) | Substitution property of equality |
| \(2m\angle B=90^\circ\) | Simplify |
| \(m\angle B=45^\circ\) | Division property of equality |
| \(\overline{AO}\cong\overline{BO}\) | Property of Circumcenter (The circumcenter is equidistant from the vertices of the triangle) |
| \(m\angle A=m\angle B\) | |
Well, this is taking some time...
Anyway, \(\triangle AOB\) is another isosceles triangle. \(m\angle OAB=m\angle B=45^\circ\), so the remaining angle, \(\angle BOA=90^\circ\)
+2446 The second one was definitely much harder for me to come up with! Here is a picture that may be useful to reference as I solve:
You will probably notice that I added a circle that inscribes the triangle. At its bost basic level, the circumcenter is a point that is equidistant from all the vertices of the triangle. The circumcenter is also the center of a triangle's circumcircle, which I have illustrated with the diagram above. Of course, we know that the sum of \(\angle AHC\) and \(\angle AOC\) is 240 by the given info. Notice that \(m\angle AOC=m\angle AHC\) because they both intercept the same arc. Therefore, we can solve for the angle of the circumcenter.
| \(m\angle AOC+m\angle AHC=240\) | Since we already established that \(m\angle AOC=m\angle AHC\), we can solve for the the central angle of the circle. |
| \(m\angle AOC+m\angle AOC=240\) | |
| \(2m\angle AOC=240\) | |
| \(m\angle AOC=120^\circ\) | There is no need to solve for the angle measure of the orthocenter; it is simply a waster of time. |
Notice that \(\angle ABC\) is an inscribed angle, which means that its measure is equal to half the measure of the central angle.
| \(\frac{1}{2}m\angle AOC=m\angle ABC\) | Use substitution here since we already know the measure of one of the angles. |
| \(\frac{1}{2}*120=m\angle ABC\) | |
| \(m\angle ABC=60^\circ\) |
+130085 1 )
The circumcenter of a right triangle always occurs at the midpoint of the hypotenuse
Thus....by SSS, triangle BOA is congruent to triangle COA
Thus angle AOB = angle AOC
But, by Euclid....a line standing upon a line making adjacent angles equal means that the lines are perpendicular.....thus angle AOB = 90°
