Do not use the thought of homothetic triangles...
Let $\triangle ABC$ and $\triangle A'B'C'$ be two non congruent triangles whose sides are relatively parallel, as in figure 1.
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Homothetic Triangles!!
This is another interesting problem relating homothetic triangles....
Do not use the thought of homothetic triangles...
Let $\triangle ABC$ and $\triangle A'B'C'$ be two non congruent triangles whose sides are relatively parallel, as in figure 1.
Then prove that the three lines $AA'$, $BB'$, $CC'$ (extended) are concurrent.
Do not use the thought of homothetic triangles...
Let $\triangle ABC$ and $\triangle A'B'C'$ be two non congruent triangles whose sides are relatively parallel, as in figure 1.
Then prove that the three lines $AA'$, $BB'$, $CC'$ (extended) are concurrent.
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Re: Homothetic Triangles!!
Hint :
Use the property no similar triangle.
Use the property no similar triangle.
One one thing is neutral in the universe, that is $0$.
Re: Homothetic Triangles!!
Masum Vaiiiiiiiiiii......
I'm STUCK!!!!
I'm STUCK!!!!
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
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tarek like math
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- Joined:Fri Feb 18, 2011 11:30 pm
Re: Homothetic Triangles!!
if A`C` extend meet C`` with BC then angle C=C``=C` bcz B`C` parallel BC and A`C` parallel AC. thus angle A=A` B=B`. so why it said not congruent=similar ?
Re: Homothetic Triangles!!
Let $O$ be the point where $AA'$ and $BB'$ meets.
Then see what $OC$ contains.
Then see what $OC$ contains.
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Nur Muhammad Shafiullah | Mahi
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Re: Homothetic Triangles!!
Tarek, in bangla,
CONGRUENT=SHORBOSHOMO;
SIMILAR=SHODRISH;
I hope you got it.
CONGRUENT=SHORBOSHOMO;
SIMILAR=SHODRISH;
I hope you got it.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Re: Homothetic Triangles!!
let $AA'$ intersects $B'C'$ and $BC$ at $D'$ and $D$.
such like,$BB'$ intersects $A'C'$ and $AC$ at $E'$ and $E$.
and ,$CC'$ intersects $A'B'$ and $AB$ at $F'$ and $F$.
look,if they are concurrent,in triangle $A'B'C'$,\[\frac{A'F'}{F'B'}\frac{B'D'}{D'C'}\frac{C'E'}{E'A'}=1\]
in triangle $ABC$,it is\[\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA}=1\]
then we get ,\[\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA}\frac{F'B'}{A'F'}\frac{D'C'}{B'D'}\frac{E'A'}{C'E'}=1.............(1)\]
dividing equation the 2nd by 1st.
it's enough to show that (1) is truth.
assume $AD$, intersects $BE,CF$ at $P,Q$.
$BE ,CF$ intersects at $R$.
now in triangle $FPB$ and $F'PB'$ for similarity(i'm not giving why they are similar,you can find it)
\[\frac{F'B'}{FB}=\frac{PB'}{PB}\]
in triangle $PB'D'$ and $PBD$ for similarity,
\[\frac{B'D'}{BD}=\frac{PB'}{PB}\]
so,\[\frac{B'D'}{BD}=\frac{F'B'}{FB}.......(2)\]
such like we can also show,\[\frac{DC}{D'C'}=\frac{CE}{C'E'}......(3)\]
\[\frac{EA}{E'A'}=\frac{AF}{A'F'}......(4)\]
from (2),(3),(4) we see (1) is truth.ceva's theorem works here.so $P,Q,R$ can't be different.
so,$AA',BB',CC'$ are concurrent.
such like,$BB'$ intersects $A'C'$ and $AC$ at $E'$ and $E$.
and ,$CC'$ intersects $A'B'$ and $AB$ at $F'$ and $F$.
look,if they are concurrent,in triangle $A'B'C'$,\[\frac{A'F'}{F'B'}\frac{B'D'}{D'C'}\frac{C'E'}{E'A'}=1\]
in triangle $ABC$,it is\[\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA}=1\]
then we get ,\[\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA}\frac{F'B'}{A'F'}\frac{D'C'}{B'D'}\frac{E'A'}{C'E'}=1.............(1)\]
dividing equation the 2nd by 1st.
it's enough to show that (1) is truth.
assume $AD$, intersects $BE,CF$ at $P,Q$.
$BE ,CF$ intersects at $R$.
now in triangle $FPB$ and $F'PB'$ for similarity(i'm not giving why they are similar,you can find it)
\[\frac{F'B'}{FB}=\frac{PB'}{PB}\]
in triangle $PB'D'$ and $PBD$ for similarity,
\[\frac{B'D'}{BD}=\frac{PB'}{PB}\]
so,\[\frac{B'D'}{BD}=\frac{F'B'}{FB}.......(2)\]
such like we can also show,\[\frac{DC}{D'C'}=\frac{CE}{C'E'}......(3)\]
\[\frac{EA}{E'A'}=\frac{AF}{A'F'}......(4)\]
from (2),(3),(4) we see (1) is truth.ceva's theorem works here.so $P,Q,R$ can't be different.
so,$AA',BB',CC'$ are concurrent.
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Re: Homothetic Triangles!!
I stuck to the problem for quite a time and solved it at last.
It seems like photon's posted a proof and it's probably the one I've used as well so I ain't posting my one. Thanks everybody.
(BTW, PHOTON, what's your actual name??)
It seems like photon's posted a proof and it's probably the one I've used as well so I ain't posting my one. Thanks everybody.
(BTW, PHOTON, what's your actual name??)
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Re: Homothetic Triangles!!
Labib wrote: (BTW, PHOTON, what's your actual name??)
This is a forum... and the name may remain hidden...
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi