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A hard Geometry Problem

Posted: Thu Feb 01, 2018 8:14 pm
by prottoydas
$\triangle ABC$ is an acute angled triangle with orthocenter $H$ and circumcenter $O$. Prove that, $\angle CAH =\angle BAO$.

Re: A hard Geometry Problem

Posted: Wed Mar 07, 2018 3:45 pm
by prottoy das
can anybody do this

Re: A hard Geometry Problem

Posted: Wed Mar 28, 2018 8:46 pm
by soyeb pervez jim
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 $\angle{CAH}=\angle{BAO}$

Re: A hard Geometry Problem

Posted: Mon Dec 03, 2018 3:10 pm
by thczarif
<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO

Re: A hard Geometry Problem

Posted: Fri Apr 10, 2020 8:18 pm
by NABILA
thczarif wrote:
Mon Dec 03, 2018 3:10 pm
<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO
Easy to understand.....