How to prove that the midpoint of the given line segment lies on another line segment?
$begingroup$
I have the following question with me:
In a triangle ABC D and E lie on sides BC and AB respectively. F lies on AC such that EF is parallel to BC, G lies on side BC such that EG lies parallel to AD. Let M and N be midpoints of AD and BC respectively. Prove that the midpoint of GF lies on line MN. Also prove that $frac{EF}{BC} + frac{EG}{AD} = 1$
How do I solve this question? I do not want to use coordinate geometry or vectors to solve this euqstion.
The only thing I figured out is that FG is also parallel to AB. Does it help?
geometry euclidean-geometry triangle
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
In a triangle ABC D and E lie on sides BC and AB respectively. F lies on AC such that EF is parallel to BC, G lies on side BC such that EG lies parallel to AD. Let M and N be midpoints of AD and BC respectively. Prove that the midpoint of GF lies on line MN. Also prove that $frac{EF}{BC} + frac{EG}{AD} = 1$
How do I solve this question? I do not want to use coordinate geometry or vectors to solve this euqstion.
The only thing I figured out is that FG is also parallel to AB. Does it help?
geometry euclidean-geometry triangle
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
In a triangle ABC D and E lie on sides BC and AB respectively. F lies on AC such that EF is parallel to BC, G lies on side BC such that EG lies parallel to AD. Let M and N be midpoints of AD and BC respectively. Prove that the midpoint of GF lies on line MN. Also prove that $frac{EF}{BC} + frac{EG}{AD} = 1$
How do I solve this question? I do not want to use coordinate geometry or vectors to solve this euqstion.
The only thing I figured out is that FG is also parallel to AB. Does it help?
geometry euclidean-geometry triangle
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
$endgroup$
I have the following question with me:
In a triangle ABC D and E lie on sides BC and AB respectively. F lies on AC such that EF is parallel to BC, G lies on side BC such that EG lies parallel to AD. Let M and N be midpoints of AD and BC respectively. Prove that the midpoint of GF lies on line MN. Also prove that $frac{EF}{BC} + frac{EG}{AD} = 1$
How do I solve this question? I do not want to use coordinate geometry or vectors to solve this euqstion.
The only thing I figured out is that FG is also parallel to AB. Does it help?
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
edited Jan 6 at 7:34
Michael Rozenberg
108k1895200
108k1895200
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
asked Jan 6 at 6:29
saisanjeevsaisanjeev
1,025312
1,025312
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $frac{AF}{AC}=frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.
Thus, $$frac{BG}{BD}=frac{BE}{AB}=1-x$$ and from here
$$vec{KN}=frac{1}{2}left(vec{FC}+vec{GB}right)=frac{1}{2}left((1-x)vec{AB}+(1-x)vec{DB}right)=frac{1}{2}(1-x)vec{MN},$$
which says $Kin MN.$
The second is true because $$frac{EF}{BC}+frac{EG}{AD}=frac{AE}{AB}+frac{BE}{AB}=1.$$
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $frac{AF}{AC}=frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.
Thus, $$frac{BG}{BD}=frac{BE}{AB}=1-x$$ and from here
$$vec{KN}=frac{1}{2}left(vec{FC}+vec{GB}right)=frac{1}{2}left((1-x)vec{AB}+(1-x)vec{DB}right)=frac{1}{2}(1-x)vec{MN},$$
which says $Kin MN.$
The second is true because $$frac{EF}{BC}+frac{EG}{AD}=frac{AE}{AB}+frac{BE}{AB}=1.$$
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
add a comment |
$begingroup$
Let $frac{AF}{AC}=frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.
Thus, $$frac{BG}{BD}=frac{BE}{AB}=1-x$$ and from here
$$vec{KN}=frac{1}{2}left(vec{FC}+vec{GB}right)=frac{1}{2}left((1-x)vec{AB}+(1-x)vec{DB}right)=frac{1}{2}(1-x)vec{MN},$$
which says $Kin MN.$
The second is true because $$frac{EF}{BC}+frac{EG}{AD}=frac{AE}{AB}+frac{BE}{AB}=1.$$
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
add a comment |
$begingroup$
Let $frac{AF}{AC}=frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.
Thus, $$frac{BG}{BD}=frac{BE}{AB}=1-x$$ and from here
$$vec{KN}=frac{1}{2}left(vec{FC}+vec{GB}right)=frac{1}{2}left((1-x)vec{AB}+(1-x)vec{DB}right)=frac{1}{2}(1-x)vec{MN},$$
which says $Kin MN.$
The second is true because $$frac{EF}{BC}+frac{EG}{AD}=frac{AE}{AB}+frac{BE}{AB}=1.$$
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Let $frac{AF}{AC}=frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.
Thus, $$frac{BG}{BD}=frac{BE}{AB}=1-x$$ and from here
$$vec{KN}=frac{1}{2}left(vec{FC}+vec{GB}right)=frac{1}{2}left((1-x)vec{AB}+(1-x)vec{DB}right)=frac{1}{2}(1-x)vec{MN},$$
which says $Kin MN.$
The second is true because $$frac{EF}{BC}+frac{EG}{AD}=frac{AE}{AB}+frac{BE}{AB}=1.$$
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 6 at 7:31
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 6:46
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
add a comment |
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
$begingroup$
thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors?
$endgroup$
– saisanjeev
Jan 6 at 9:53
add a comment |
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