cool geo

For discussing Olympiad level Geometry Problems
Tahmid
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cool geo

Unread post by Tahmid » Thu Apr 23, 2015 11:29 am

In $\Delta ABC$, $AD,BE,CF$ are the feets of the perpendiculars . $M,N$ are the midpoints of $BC$ and $AH$ respectively [$H$ orthocentre] . $MN$ intersects the internal and the external angle bisector of $\angle BAC$ at $P$ and $Q$ respectively. Prove that $P,Q$ lies on the circumcircle of $\Delta AEF$

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sowmitra
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Location: Mirpur, Dhaka, Bangladesh

Re: cool geo

Unread post by sowmitra » Sun Apr 26, 2015 10:31 pm

$AEHF$ is cyclic. $\therefore P \in \odot AEF$ iff $\angle APH= \angle AFH =90^{\circ}$.

Suppose, $AP\cap\odot ABC=G$.
Since, $G$ is the mid-point of arc $\widehat{BC}$, $M$ is the mid-point of $OG$.

Now, $AN||OM$ and $AN=OM$ $\Rightarrow AOMN$ is a prallelogram
$\Rightarrow AO||MN$ $\Rightarrow \triangle GMP \sim \triangle GOA$.
But, $OG=OA$. $\therefore MP=MG$.
$\Rightarrow \angle MGP=\angle MPG \Rightarrow \angle NAP=\angle NPA \Rightarrow NA=NP$.
But, $NA=NH$. So, $N$ is the centre of $\odot AHP$ and $\angle AHP=90^{\circ}$.
$\therefore P$ lies on $\odot AEF$. Similarly, $Q \in \odot AEF$.
Last edited by sowmitra on Mon Apr 27, 2015 7:40 pm, edited 1 time in total.
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: cool geo

Unread post by Tahmid » Mon Apr 27, 2015 10:37 am

sowmitra wrote:$AEHF$ is cyclic. $\therefore P \in \odot AEF$ iff $\angle AHP= \angle AHF =90^{\circ}$.
here is a little typo . $\angle APH= \angle AFH$ should be $90$

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sowmitra
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Re: cool geo

Unread post by sowmitra » Mon Apr 27, 2015 7:41 pm

Corrected. :roll:
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: cool geo

Unread post by Tahmid » Tue Apr 28, 2015 3:22 am

vaia your solution is very nice .
but i did it just using angle chasing . :)

let the internal angle bisector intersects the side $BC$ at $X$
$\angle NAP = 90-\angle AXD=90- \angle A/2 -\angle C$

$\angle NPA=\angle XPM=180-\angle PMX-\angle PXM=180-(180-\angle NFD)-\angle A/2-\angle B$
$=180-180+\angle B+90-\angle C-\angle A/2-\angle B=90- \angle A/2 -\angle C$

so, $\angle NAP=\angle NPA$
$\therefore NP=NA$
that means $P$ lies on the circumcircle of $AEF$

similarly for $Q$

so, we are done !!!

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