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Might be slightly generalizable
Rijul saini   5
N an hour ago by guptaamitu1
Source: India IMOTC Day 3 Problem 1
Let $ABC$ be an acute angled triangle with orthocenter $H$ and $AB<AC$. Let $T(\ne B,C, H)$ be any other point on the arc $\stackrel{\LARGE\frown}{BHC}$ of the circumcircle of $BHC$ and let line $BT$ intersect line $AC$ at $E(\ne A)$ and let line $CT$ intersect line $AB$ at $F(\ne A)$. Let the circumcircles of $AEF$ and $ABC$ intersect again at $X$ ($\ne A$). Let the lines $XE,XF,XT$ intersect the circumcircle of $(ABC)$ again at $P,Q,R$ ($\ne X$). Prove that the lines $AR,BC,PQ$ concur.
5 replies
Rijul saini
Yesterday at 6:39 PM
guptaamitu1
an hour ago
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Incircle touch points are on circle
zuss77   3
N Jul 16, 2018 by jayme
Source: Kyiv Geometry O.
$\triangle ABC$. $\angle A=90^{\circ}$.
Incircle touches $AB$ and $AC$ at $D, E$.
Middle line parallel to $BC$ cuts $\odot(ABC)$ at $P, T$.
Prove that $D, E, P, T$ are lay on circle.

3 replies
zuss77
Jul 15, 2018
jayme
Jul 16, 2018
Incircle touch points are on circle
G H J
Source: Kyiv Geometry O.
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zuss77
520 posts
#1 • 2 Y
Y by Adventure10, Mango247
$\triangle ABC$. $\angle A=90^{\circ}$.
Incircle touches $AB$ and $AC$ at $D, E$.
Middle line parallel to $BC$ cuts $\odot(ABC)$ at $P, T$.
Prove that $D, E, P, T$ are lay on circle.
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buratinogigle
2405 posts
#2 • 2 Y
Y by Adventure10, Mango247
See https://ijgeometry.com/product/tran-quang-hung-feuerbachs-theorem-right-triangle-extension/
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zuss77
520 posts
#3 • 2 Y
Y by Adventure10, Mango247
Another solution:

Let $Q=DE \cap PT$.
Let $R=AQ \cap BC$.
Let $AQ$ cut $\odot(ABC)$ second time at $K$.

1) $RI \perp AI$ >>

2) $IK \perp AR$ >>

3) $\odot (DEPT)$ >>
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jayme
9805 posts
#4 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

see : http://jl.ayme.pagesperso-orange.fr/Docs/Quelques%20theoremes%20oublies.pdf

Sincerely
Jean-Louis
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