Triangle centers Orthocenter

kaloyan

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Apr 27, 2019
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An acute [MATH]\triangle ABC[/MATH], inscribed in a circle [MATH]k[/MATH] with radii [MATH]R[/MATH], is given. Point [MATH]H[/MATH] is the orthocenter of [MATH]\triangle ABC[/MATH] and [MATH]AH=R[/MATH]. Find [MATH]\angle BAC[/MATH]. (Answer: [MATH]60^\circ[/MATH])
12019
[MATH]AD[/MATH] is diameter, thus [MATH]\angle ACD = \angle ABD = 90^\circ[/MATH]. Also [MATH]HBDC[/MATH] is parallelogram because [MATH]HC || BD, HB||CD[/MATH]. It seems useless and I don't know how to continue. Thank you in advance!
 
An acute [MATH]\triangle ABC[/MATH], inscribed in a circle [MATH]k[/MATH] with radii [MATH]R[/MATH], is given. Point [MATH]H[/MATH] is the orthocenter of [MATH]\triangle ABC[/MATH] and [MATH]AH=R[/MATH]. Find [MATH]\angle BAC[/MATH]. (Answer: [MATH]60^\circ[/MATH])
View attachment 12019
[MATH]AD[/MATH] is diameter, thus [MATH]\angle ACD = \angle ABD = 90^\circ[/MATH]. Also [MATH]HBDC[/MATH] is parallelogram because [MATH]HC || BD, HB||CD[/MATH]. It seems useless and I don't know how to continue. Thank you in advance!
You have to try. Try labeling some angles and see what happens! Show us your updated diagram and if you still need some help you will be given a hint.
 
You have to try. Try labeling some angles and see what happens! Show us your updated diagram and if you still need some help you will be given a hint.
Thank you for your response! Under the diagram was my first try. I've already solved the problem.
 
I asked you in another forum what graphics program you used for that figure. Care to answer? Also, it would be good form for you to post your solution now that you have solved it.
 
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