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Proof for similarities between two triangles.
The 2019 Stack Overflow Developer Survey Results Are InWhy are there isosceles triangles?Congruent parts of trianglesAre there two triangles with equal angles and two equal sides which are not congruent?Is it possible that two triangles satisfy these conditions?How to proof that those triangles are similar?How is it proved that 3 inradii multiplied by 3 inradii of similar triangles are equal?Similar Triangles Problem (No corresponding sides)Is it possible to have two triangles with equal sides but with different angles?congruence of two trianglesProblem with 2 similar right-angled triangles
$begingroup$
We know that if the angles of two triangles are similar, then their sides are proportional. I get the idea. Now, can it be proven rigorously?
euclidean-geometry triangles formal-proofs
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We know that if the angles of two triangles are similar, then their sides are proportional. I get the idea. Now, can it be proven rigorously?
euclidean-geometry triangles formal-proofs
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10
add a comment |
$begingroup$
We know that if the angles of two triangles are similar, then their sides are proportional. I get the idea. Now, can it be proven rigorously?
euclidean-geometry triangles formal-proofs
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know that if the angles of two triangles are similar, then their sides are proportional. I get the idea. Now, can it be proven rigorously?
euclidean-geometry triangles formal-proofs
euclidean-geometry triangles formal-proofs
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 20:01
LikhonLikhon
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 20:01
LikhonLikhon
81
asked Apr 6 at 20:01
LikhonLikhon
81
81
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Likhon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10
add a comment |
$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10
$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Through a rigid motion, you can have angles $widehatBAC$ and $widehatB'A'C'$ coincide with additional conditions $A'=A$, $B'in l(AB)$ and $C'in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'parallel BC$. Consider now $rparallel l(BC)$ such that $A=A'in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$
Now that you know this proportionality, the two triangles $triangle ABC$ and $triangle A'B'C'$ fall into the SAS criterion for proportionality.
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
$endgroup$
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
add a comment |
Your Answer
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1 Answer
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$begingroup$
Through a rigid motion, you can have angles $widehatBAC$ and $widehatB'A'C'$ coincide with additional conditions $A'=A$, $B'in l(AB)$ and $C'in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'parallel BC$. Consider now $rparallel l(BC)$ such that $A=A'in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$
Now that you know this proportionality, the two triangles $triangle ABC$ and $triangle A'B'C'$ fall into the SAS criterion for proportionality.
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
$endgroup$
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
add a comment |
$begingroup$
Through a rigid motion, you can have angles $widehatBAC$ and $widehatB'A'C'$ coincide with additional conditions $A'=A$, $B'in l(AB)$ and $C'in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'parallel BC$. Consider now $rparallel l(BC)$ such that $A=A'in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$
Now that you know this proportionality, the two triangles $triangle ABC$ and $triangle A'B'C'$ fall into the SAS criterion for proportionality.
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
$endgroup$
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
add a comment |
$begingroup$
Through a rigid motion, you can have angles $widehatBAC$ and $widehatB'A'C'$ coincide with additional conditions $A'=A$, $B'in l(AB)$ and $C'in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'parallel BC$. Consider now $rparallel l(BC)$ such that $A=A'in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$
Now that you know this proportionality, the two triangles $triangle ABC$ and $triangle A'B'C'$ fall into the SAS criterion for proportionality.
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
$endgroup$
Through a rigid motion, you can have angles $widehatBAC$ and $widehatB'A'C'$ coincide with additional conditions $A'=A$, $B'in l(AB)$ and $C'in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'parallel BC$. Consider now $rparallel l(BC)$ such that $A=A'in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$
Now that you know this proportionality, the two triangles $triangle ABC$ and $triangle A'B'C'$ fall into the SAS criterion for proportionality.
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
answered Apr 6 at 20:25
Saucy O'PathSaucy O'Path
6,5581627
6,5581627
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
add a comment |
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
$begingroup$
Thanks, that thales theorem was the missing piece for me.
$endgroup$
– Likhon
Apr 6 at 20:37
add a comment |
Likhon is a new contributor. Be nice, and check out our Code of Conduct.
Likhon is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Do you know law of sines? If you don't see here mathsisfun.com/algebra/trig-sine-law.html
$endgroup$
– Dbchatto67
Apr 6 at 20:06
$begingroup$
Yes I know that. But doesn't the proof that, "sin/cos... of an angle is equal for all right triangles" come from this? If that does then I think we shouldn't be using trigonometry.
$endgroup$
– Likhon
Apr 6 at 20:10