Geometry - Proving a common centroid.












2












$begingroup$



Triangle PQR is drawn. Through it's vertices are lines drawn which are
parallel to the opposite sides of the triangle. The new triangle formed
is ABC. Prove that these two triangles have a common centroid.




I started by letting $M$ be the median of $[PQ]$, and then prove that bisector of $[BC]$ from $A$ occurs at $R$ while $M$ is collinear to $[AP]$. Firstly, I'm not sure whether or not this will suffice in proving they share a common centroid, and secondly I'm not sure where to start with the proof as setting up similar triangles from the parallel lines identity has lead to nothing.








geometry euclidean-geometry







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$endgroup$












  • $begingroup$
    $exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
    $endgroup$
    – Shamim Akhtar
    42 mins ago










  • $begingroup$
    I am not talking about similarity, homothety
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    You dont know homothety?
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    So what if you are in high school? I am in middle school
    $endgroup$
    – Shamim Akhtar
    33 mins ago










  • $begingroup$
    Okay i am trying to help you without homothety
    $endgroup$
    – Shamim Akhtar
    33 mins ago
















2












$begingroup$



Triangle PQR is drawn. Through it's vertices are lines drawn which are
parallel to the opposite sides of the triangle. The new triangle formed
is ABC. Prove that these two triangles have a common centroid.




I started by letting $M$ be the median of $[PQ]$, and then prove that bisector of $[BC]$ from $A$ occurs at $R$ while $M$ is collinear to $[AP]$. Firstly, I'm not sure whether or not this will suffice in proving they share a common centroid, and secondly I'm not sure where to start with the proof as setting up similar triangles from the parallel lines identity has lead to nothing.








geometry euclidean-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    $exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
    $endgroup$
    – Shamim Akhtar
    42 mins ago










  • $begingroup$
    I am not talking about similarity, homothety
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    You dont know homothety?
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    So what if you are in high school? I am in middle school
    $endgroup$
    – Shamim Akhtar
    33 mins ago










  • $begingroup$
    Okay i am trying to help you without homothety
    $endgroup$
    – Shamim Akhtar
    33 mins ago














2












2








2





$begingroup$



Triangle PQR is drawn. Through it's vertices are lines drawn which are
parallel to the opposite sides of the triangle. The new triangle formed
is ABC. Prove that these two triangles have a common centroid.




I started by letting $M$ be the median of $[PQ]$, and then prove that bisector of $[BC]$ from $A$ occurs at $R$ while $M$ is collinear to $[AP]$. Firstly, I'm not sure whether or not this will suffice in proving they share a common centroid, and secondly I'm not sure where to start with the proof as setting up similar triangles from the parallel lines identity has lead to nothing.








geometry euclidean-geometry







share|cite|improve this question









$endgroup$





Triangle PQR is drawn. Through it's vertices are lines drawn which are
parallel to the opposite sides of the triangle. The new triangle formed
is ABC. Prove that these two triangles have a common centroid.




I started by letting $M$ be the median of $[PQ]$, and then prove that bisector of $[BC]$ from $A$ occurs at $R$ while $M$ is collinear to $[AP]$. Firstly, I'm not sure whether or not this will suffice in proving they share a common centroid, and secondly I'm not sure where to start with the proof as setting up similar triangles from the parallel lines identity has lead to nothing.








geometry euclidean-geometry




geometry euclidean-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 53 mins ago









John MillerJohn Miller

1747




1747












  • $begingroup$
    $exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
    $endgroup$
    – Shamim Akhtar
    42 mins ago










  • $begingroup$
    I am not talking about similarity, homothety
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    You dont know homothety?
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    So what if you are in high school? I am in middle school
    $endgroup$
    – Shamim Akhtar
    33 mins ago










  • $begingroup$
    Okay i am trying to help you without homothety
    $endgroup$
    – Shamim Akhtar
    33 mins ago


















  • $begingroup$
    $exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
    $endgroup$
    – Shamim Akhtar
    42 mins ago










  • $begingroup$
    I am not talking about similarity, homothety
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    You dont know homothety?
    $endgroup$
    – Shamim Akhtar
    38 mins ago










  • $begingroup$
    So what if you are in high school? I am in middle school
    $endgroup$
    – Shamim Akhtar
    33 mins ago










  • $begingroup$
    Okay i am trying to help you without homothety
    $endgroup$
    – Shamim Akhtar
    33 mins ago
















$begingroup$
$exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
$endgroup$
– Shamim Akhtar
42 mins ago




$begingroup$
$exists$ a homothety mapping the sides of $Delta PQR$ to the sides of $Delta ABC$.
$endgroup$
– Shamim Akhtar
42 mins ago












$begingroup$
I am not talking about similarity, homothety
$endgroup$
– Shamim Akhtar
38 mins ago




$begingroup$
I am not talking about similarity, homothety
$endgroup$
– Shamim Akhtar
38 mins ago












$begingroup$
You dont know homothety?
$endgroup$
– Shamim Akhtar
38 mins ago




$begingroup$
You dont know homothety?
$endgroup$
– Shamim Akhtar
38 mins ago












$begingroup$
So what if you are in high school? I am in middle school
$endgroup$
– Shamim Akhtar
33 mins ago




$begingroup$
So what if you are in high school? I am in middle school
$endgroup$
– Shamim Akhtar
33 mins ago












$begingroup$
Okay i am trying to help you without homothety
$endgroup$
– Shamim Akhtar
33 mins ago




$begingroup$
Okay i am trying to help you without homothety
$endgroup$
– Shamim Akhtar
33 mins ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.



$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.



It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.



Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
    $endgroup$
    – John Miller
    20 mins ago





















2












$begingroup$

Solution with vectors. The centroid $G$ of $ABC$ is $G = {1over 3}(A+B+C)$



Since $R$ is a midpoint of $BC$ we have $$R = {1over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1over 3}(P+Q+R) = {1over 3}Big({1over 2}(B+C)+{1over 2}(A+C)+ {1over 2}(B+A)Big) $$$$= {1over 3}(A+B+C) = G$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, I do feel as if the proof using a parallelogram is sufficient.
    $endgroup$
    – John Miller
    6 mins ago



















1












$begingroup$

Notice that $angle CQR=angle QRP=angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I'm not sure which answer to give the check to as they both answer the question.
    $endgroup$
    – John Miller
    20 mins ago










  • $begingroup$
    You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
    $endgroup$
    – Shamim Akhtar
    14 mins ago










  • $begingroup$
    Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
    $endgroup$
    – John Miller
    8 mins ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.



$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.



It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.



Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
    $endgroup$
    – John Miller
    20 mins ago


















2












$begingroup$

$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.



$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.



It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.



Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
    $endgroup$
    – John Miller
    20 mins ago
















2












2








2





$begingroup$

$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.



$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.



It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.



Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.






share|cite|improve this answer









$endgroup$



$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.



$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.



It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.



Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 25 mins ago









AretinoAretino

26.1k31546




26.1k31546












  • $begingroup$
    +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
    $endgroup$
    – John Miller
    20 mins ago




















  • $begingroup$
    +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
    $endgroup$
    – John Miller
    20 mins ago


















$begingroup$
+1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
$endgroup$
– John Miller
20 mins ago






$begingroup$
+1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too.
$endgroup$
– John Miller
20 mins ago













2












$begingroup$

Solution with vectors. The centroid $G$ of $ABC$ is $G = {1over 3}(A+B+C)$



Since $R$ is a midpoint of $BC$ we have $$R = {1over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1over 3}(P+Q+R) = {1over 3}Big({1over 2}(B+C)+{1over 2}(A+C)+ {1over 2}(B+A)Big) $$$$= {1over 3}(A+B+C) = G$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, I do feel as if the proof using a parallelogram is sufficient.
    $endgroup$
    – John Miller
    6 mins ago
















2












$begingroup$

Solution with vectors. The centroid $G$ of $ABC$ is $G = {1over 3}(A+B+C)$



Since $R$ is a midpoint of $BC$ we have $$R = {1over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1over 3}(P+Q+R) = {1over 3}Big({1over 2}(B+C)+{1over 2}(A+C)+ {1over 2}(B+A)Big) $$$$= {1over 3}(A+B+C) = G$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, I do feel as if the proof using a parallelogram is sufficient.
    $endgroup$
    – John Miller
    6 mins ago














2












2








2





$begingroup$

Solution with vectors. The centroid $G$ of $ABC$ is $G = {1over 3}(A+B+C)$



Since $R$ is a midpoint of $BC$ we have $$R = {1over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1over 3}(P+Q+R) = {1over 3}Big({1over 2}(B+C)+{1over 2}(A+C)+ {1over 2}(B+A)Big) $$$$= {1over 3}(A+B+C) = G$$






share|cite|improve this answer











$endgroup$



Solution with vectors. The centroid $G$ of $ABC$ is $G = {1over 3}(A+B+C)$



Since $R$ is a midpoint of $BC$ we have $$R = {1over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1over 3}(P+Q+R) = {1over 3}Big({1over 2}(B+C)+{1over 2}(A+C)+ {1over 2}(B+A)Big) $$$$= {1over 3}(A+B+C) = G$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 7 mins ago

























answered 13 mins ago









Maria MazurMaria Mazur

51k1362126




51k1362126








  • 1




    $begingroup$
    Thank you, I do feel as if the proof using a parallelogram is sufficient.
    $endgroup$
    – John Miller
    6 mins ago














  • 1




    $begingroup$
    Thank you, I do feel as if the proof using a parallelogram is sufficient.
    $endgroup$
    – John Miller
    6 mins ago








1




1




$begingroup$
Thank you, I do feel as if the proof using a parallelogram is sufficient.
$endgroup$
– John Miller
6 mins ago




$begingroup$
Thank you, I do feel as if the proof using a parallelogram is sufficient.
$endgroup$
– John Miller
6 mins ago











1












$begingroup$

Notice that $angle CQR=angle QRP=angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I'm not sure which answer to give the check to as they both answer the question.
    $endgroup$
    – John Miller
    20 mins ago










  • $begingroup$
    You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
    $endgroup$
    – Shamim Akhtar
    14 mins ago










  • $begingroup$
    Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
    $endgroup$
    – John Miller
    8 mins ago
















1












$begingroup$

Notice that $angle CQR=angle QRP=angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I'm not sure which answer to give the check to as they both answer the question.
    $endgroup$
    – John Miller
    20 mins ago










  • $begingroup$
    You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
    $endgroup$
    – Shamim Akhtar
    14 mins ago










  • $begingroup$
    Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
    $endgroup$
    – John Miller
    8 mins ago














1












1








1





$begingroup$

Notice that $angle CQR=angle QRP=angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.






share|cite|improve this answer









$endgroup$



Notice that $angle CQR=angle QRP=angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 24 mins ago









Shamim AkhtarShamim Akhtar

59319




59319












  • $begingroup$
    Thank you, I'm not sure which answer to give the check to as they both answer the question.
    $endgroup$
    – John Miller
    20 mins ago










  • $begingroup$
    You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
    $endgroup$
    – Shamim Akhtar
    14 mins ago










  • $begingroup$
    Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
    $endgroup$
    – John Miller
    8 mins ago


















  • $begingroup$
    Thank you, I'm not sure which answer to give the check to as they both answer the question.
    $endgroup$
    – John Miller
    20 mins ago










  • $begingroup$
    You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
    $endgroup$
    – Shamim Akhtar
    14 mins ago










  • $begingroup$
    Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
    $endgroup$
    – John Miller
    8 mins ago
















$begingroup$
Thank you, I'm not sure which answer to give the check to as they both answer the question.
$endgroup$
– John Miller
20 mins ago




$begingroup$
Thank you, I'm not sure which answer to give the check to as they both answer the question.
$endgroup$
– John Miller
20 mins ago












$begingroup$
You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
$endgroup$
– Shamim Akhtar
14 mins ago




$begingroup$
You can check the other answer, it is obviouslu better than mine, it uses elementary techniques.
$endgroup$
– Shamim Akhtar
14 mins ago












$begingroup$
Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
$endgroup$
– John Miller
8 mins ago




$begingroup$
Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple.
$endgroup$
– John Miller
8 mins ago


















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Ponta tanko

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Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
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Tantalo (mitologio)

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Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
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Erzsébet Schaár

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Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
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