Find this angle given the triangle's Orthocenter

[kaloyan]

Homework Statement

An acute $\triangle ABC$, inscribed in a circle $k$ with radii $R$, is given. Point $H$ is the orthocenter of $\triangle ABC$ and $AH=R$. Find $\angle BAC$. (Answer: $60^\circ$)

Relevant Equations

-

$AD$ is diameter, thus $\angle ACD = \angle ABD = 90^\circ$. Also $HBDC$ is a parallelogram because $HC||BD, HB||CD$. It seems useless and I don't know how to continue. Thank you in advance!

 

[LCKurtz]

What drawing program did you use to make that figure?

 

[kaloyan]

What drawing program did you use to make that figure?

I've used GeoGebra.

 

[ali PMPAINT]

This actually doesn't have one answer. just put A=70 for example, and it won't go wrong, or 60, and etc.

 

[LCKurtz]

This actually doesn't have one answer. just put A=70 for example, and it won't go wrong, or 60, and etc.

I don't get what you are trying to say. If you try to draw the same picture with an angle of $70^\circ$ you don't get the lengths $AH = AO$.

 

[ali PMPAINT]

I don't get what you are trying to say. If you try to draw the same picture with an angle of $70^\circ$ you don't get the lengths $AH = AO$.

Oh, yes, you are right. For reasons, I thought any angle would be correct

 

[ali PMPAINT]

Problem Statement: An acute $\triangle ABC$, inscribed in a circle $k$ with radii $R$, is given. Point $H$ is the orthocenter of $\triangle ABC$ and $AH=R$. Find $\angle BAC$. (Answer: $60^\circ$)
Relevant Equations: -

View attachment 243002
$AD$ is diameter, thus $\angle ACD = \angle ABD = 90^\circ$. Also $HBDC$ is a parallelogram because $HC||BD, HB||CD$. It seems useless and I don't know how to continue. Thank you in advance!

So, you can continue by showing first portion A is equal to the third portion A, Then try to use trigonometry since AH=AO=R and AD=2R, and then you will get the answer.

 

[kaloyan]

So, you can continue by showing first portion A is equal to the third portion A, Then try to use trigonometry since AH=AO=R and AD=2R, and then you will get the answer.

I'm 8th grade - I do not know Trigonometry.

 

[ali PMPAINT]

I'm 8th grade - I do not know Trigonometry.

What about similarity? It can be solved by it.(I don't know how is your country's education system)
And another clue to solve the problem: After you showed that the first portion A is equal to the third portion A, then by this and by knowing that AH=AO=R and AD=2R , try to prove 2AE=AB (by similarity)