A hard Geometry Problem
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- Joined:Thu Feb 01, 2018 11:56 am
$\triangle ABC$ is an acute angled triangle with orthocenter $H$ and circumcenter $O$. Prove that, $\angle CAH =\angle BAO$.
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- Location:Sylhet
Re: A hard Geometry Problem
can anybody do this
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- Joined:Sat Jan 28, 2017 11:06 pm
Re: A hard Geometry Problem
Draw perpendicular from $O$ to $AB$
In $\triangle ABC$
$2\angle{ACB}=\angle{AOB}$
Again, in $\triangle {AOD}$ and $\triangle {DOB}$,
$AO=OB$, and $\angle {ODA}=\angle{ODB}=90$ so $OD$ is angular bisector of $\angle{AOB}$
so $\angle{AOD}=\angle{ACB}$
so there complimentary angles are also same.
so $\angle{CAH}=\angle{BAO}$
In $\triangle ABC$
$2\angle{ACB}=\angle{AOB}$
Again, in $\triangle {AOD}$ and $\triangle {DOB}$,
$AO=OB$, and $\angle {ODA}=\angle{ODB}=90$ so $OD$ is angular bisector of $\angle{AOB}$
so $\angle{AOD}=\angle{ACB}$
so there complimentary angles are also same.
so $\angle{CAH}=\angle{BAO}$
Re: A hard Geometry Problem
<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO