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

Unread post by prottoydas » Thu Feb 01, 2018 8:14 pm

$\triangle ABC$ 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

Unread post by prottoy das » Wed Mar 07, 2018 3:45 pm

can anybody do this

soyeb pervez jim
Posts: 21
Joined: Sat Jan 28, 2017 11:06 pm

Re: A hard Geometry Problem

Unread post by soyeb pervez jim » Wed Mar 28, 2018 8:46 pm

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

Unread post by thczarif » Mon Dec 03, 2018 3:10 pm

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

Post Reply