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Angles TXA and BAC are equal
nAalniaOMliO   1
N 5 minutes ago by Retemoeg
Source: Belarusian National Olympiad 2025
Altitudes $BE$ and $CF$ of triangle $ABC$ intersect in $H$. A perpendicular $HT$ from $H$ to $EF$ is drawn. Circumcircles $ABC$ and $BHT$ intersect at $B$ and $X$.
Prove that $\angle TXA= \angle BAC$.
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nAalniaOMliO
Friday at 8:23 PM
Retemoeg
5 minutes ago
GOTEEM #1: Incircle Concurrency
tworigami   30
N 6 minutes ago by ErTeeEs06
Source: GOTEEM: Mock Geometry Olympiad
Let $ABC$ be a scalene triangle. The incircle of $\triangle ABC$ is tangent to sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D, E, F$, respectively. Let $G$ be a point on the incircle of $\triangle ABC$ such that $\angle AGD = 90^\circ$. If lines $DG$ and $EF$ intersect at $P$, prove that $AP$ is parallel to $BC$.

Proposed by bluek
30 replies
tworigami
Jan 2, 2020
ErTeeEs06
6 minutes ago
A geometry problem
Lttgeometry   0
38 minutes ago
Given a non-isosceles triangle $ABC$ that is inscribed in $(O)$ . The incircle $(I)$ is tangent to $BC,CA,AB$ at $D,E,F$ respectively. A line through $A$ parallel to $BC$ intersects $(O)$ at $T$, and $TD$ intersects $(O)$ again at $J$. Let $N$ is the midpoint of $BC$. $P,Q$ be the second intersection of $JE,JF$ with $(O)$. $AI$ intersects $(O)$ again at $M$. Prove that the line passing through $A$ perpendicular to $PQ$ bisects $MN$.
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+2 w
Lttgeometry
38 minutes ago
0 replies
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Amazing concurency
nikolinv   0
Jul 12, 2016
Source: Original
Suprizing!
Midline, symmedian line, orthic line (side of the orthic triangle) and antiparallel through vertex are concurent.

My solution
0 replies
nikolinv
Jul 12, 2016
0 replies
Amazing concurency
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Source: Original
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nikolinv
43 posts
#1 • 1 Y
Y by Adventure10
Suprizing!
Midline, symmedian line, orthic line (side of the orthic triangle) and antiparallel through vertex are concurent.

My solution
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