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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?

 Feb 8, 2018

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!


\(\triangle ABC\text{ is an isosceles right triangle}\\ m\angle A=90^{\circ}\\ O\text{ is the circumcenter}\) 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, \(\angle BOA=90^\circ\)

 Feb 9, 2018

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.
\(\frac{1}{2}*120=m\angle ABC\) 
\(m\angle ABC=60^\circ\) 
 Feb 9, 2018

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°




cool cool cool

 Feb 9, 2018

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