## A hard Geometry Problem

For discussing Olympiad level Geometry Problems
prottoydas
Posts: 8
Joined: Thu Feb 01, 2018 11:56 am

### A hard Geometry Problem

\$\triangle ABC\$ is an acute angled triangle with orthocenter \$H\$ and circumcenter \$O\$. Prove that, \$\angle CAH =\angle BAO\$.

prottoy das
Posts: 17
Joined: Thu Feb 01, 2018 11:28 am
Location: Sylhet

### Re: A hard Geometry Problem

can anybody do this

soyeb pervez jim
Posts: 21
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}\$

thczarif
Posts: 14
Joined: Mon Sep 25, 2017 11:27 pm

### Re: A hard Geometry Problem

<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO