Show that two segments are perpendicular
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Let $ABC$ be a triangle such that $|AC|=|BC|.$ Let $M$ be the midpoint of $AB$ and let $D$ be the midpoint of $MC$. Let $S$ be the the point obtained from projecting $M$ orthogonally onto $AD.$ Show that $BS$ and $CS$ are perpendicular.
My idea: Show that you can inscribe a quadrilateral $CSMB$ in a circle. That way the angle $CSB$ equals the angle $CMB.$ To show that you you can inscribe $CSMB$ in a circle, it is enough to show one of the following:
- the angle $SCM$ equals the angle $SBM$
- The angle $MSB$ equals the angle $MCB$
- The angle $MCB$ equals the angle $MSB$
Any help would be greatly appreciated.
geometry plane-geometry
asked Aug 8 at 16:14
Pawel
2,974921
add a comment |Â
up vote
0
down vote
favorite
Let $ABC$ be a triangle such that $|AC|=|BC|.$ Let $M$ be the midpoint of $AB$ and let $D$ be the midpoint of $MC$. Let $S$ be the the point obtained from projecting $M$ orthogonally onto $AD.$ Show that $BS$ and $CS$ are perpendicular.
My idea: Show that you can inscribe a quadrilateral $CSMB$ in a circle. That way the angle $CSB$ equals the angle $CMB.$ To show that you you can inscribe $CSMB$ in a circle, it is enough to show one of the following:
- the angle $SCM$ equals the angle $SBM$
- The angle $MSB$ equals the angle $MCB$
- The angle $MCB$ equals the angle $MSB$
Any help would be greatly appreciated.
geometry plane-geometry
asked Aug 8 at 16:14
Pawel
2,974921
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $ABC$ be a triangle such that $|AC|=|BC|.$ Let $M$ be the midpoint of $AB$ and let $D$ be the midpoint of $MC$. Let $S$ be the the point obtained from projecting $M$ orthogonally onto $AD.$ Show that $BS$ and $CS$ are perpendicular.
My idea: Show that you can inscribe a quadrilateral $CSMB$ in a circle. That way the angle $CSB$ equals the angle $CMB.$ To show that you you can inscribe $CSMB$ in a circle, it is enough to show one of the following:
- the angle $SCM$ equals the angle $SBM$
- The angle $MSB$ equals the angle $MCB$
- The angle $MCB$ equals the angle $MSB$
Any help would be greatly appreciated.
geometry plane-geometry
asked Aug 8 at 16:14
Pawel
2,974921
Let $ABC$ be a triangle such that $|AC|=|BC|.$ Let $M$ be the midpoint of $AB$ and let $D$ be the midpoint of $MC$. Let $S$ be the the point obtained from projecting $M$ orthogonally onto $AD.$ Show that $BS$ and $CS$ are perpendicular.
My idea: Show that you can inscribe a quadrilateral $CSMB$ in a circle. That way the angle $CSB$ equals the angle $CMB.$ To show that you you can inscribe $CSMB$ in a circle, it is enough to show one of the following:
- the angle $SCM$ equals the angle $SBM$
- The angle $MSB$ equals the angle $MCB$
- The angle $MCB$ equals the angle $MSB$
Any help would be greatly appreciated.
geometry plane-geometry
asked Aug 8 at 16:14
Pawel
2,974921
asked Aug 8 at 16:14
Pawel
2,974921
asked Aug 8 at 16:14
Pawel
2,974921
asked Aug 8 at 16:14
Pawel
2,974921
2,974921
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
answered Aug 8 at 17:40
Michael Behrend
89225
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
answered Aug 8 at 17:40
Michael Behrend
89225
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
add a comment |Â
up vote
1
down vote
accepted
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
answered Aug 8 at 17:40
Michael Behrend
89225
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
answered Aug 8 at 17:40
Michael Behrend
89225
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
answered Aug 8 at 17:40
Michael Behrend
89225
answered Aug 8 at 17:40
Michael Behrend
89225
answered Aug 8 at 17:40
Michael Behrend
89225
answered Aug 8 at 17:40
Michael Behrend
89225
89225
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
add a comment |Â
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
How is it easy to see that $SMB=SDC?$
– Pawel
Aug 8 at 17:42
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
Oh, I see it! Thanks!
– Pawel
Aug 8 at 17:53
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
For those of you who are still not sure why, draw a line that goes through D and is parallel to AB
– Pawel
Aug 8 at 17:54
add a comment |Â
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