In triangle $ABC$: $H_{1}$ is a foot of an altitude from side $BC$, $H_{2}$ is a foot of an altitude from side $AC$, $H_{3}$ is a foot of an altitude from side $AB$, $M_{1}$ is midpoint of $BC$, $M_{2}$ is midpoint of $AC$, $M_{3}$ is midpoint of $AB$, $H$ is an orthocenter of $ABC$.
$X_{2}$ and $X_{3}$ are points that are symmetrical to $H_{1}$ relatively $BH_{2}$ and $CH_{3}$ respectively. Lines $M_{3}X_{2}$ and $M_{2}X_{3}$ intersect in point $X$. Analogically $Y_{3}$ and $Y_{1}$ are points symmetrical to $H_{2}$ relatively to $CH_{3}$ and $AH_{1}$; Lines $M_{1}Y_{3}$ and $M_{3}Y_{1}$ intersect in point $Y$. Finally, $Z_{1}$, $Z_{2}$ are points, symmetrical to $H_{3}$ relatively to $AH_{1}$ and $BH_{2}$; Lines $M_{1}Z_{2}$ and $M_{2}Z_{1}$ intersect in point Z.
Prove, that point H is incentre of triangle $XYZ$
2026-04-01 20:09:01.1775074141
Ortocenter and incenter
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Here is a link to an interactive diagram in which You can drag point B and observe that the theorem as stated only holds if the original triangle is acute angled. If it is obtuse, then it would appear that the orthocenter of ABC is one of the excenters of XYZ.
If you like automated proofs (OK, I agree they are not as good as the real thing..) here is a Geometry Expressions "proof"
By inspection (and assuming all the assumptions hold), the distance between H and the three sides of the triangle XYZ are identical algebraic expressions