Geometry problem (from national competition)
$begingroup$
Given right angled triangle $triangle ABC$ with right angle at point C.
Let points $D, E$ lie on $AB$, such that $|BC|=|BD|$ and $|AC|=|AE|$.
Let point $F$ be orthogonal projection of point $D$ onto $AC$ and let point $G$ be orthogonal projection of point $E$ onto $BC$.
Prove $|DE|=|DF|+|EG|$.
What I've got so far:
EDIT: Had a picture here, but I removed it because it was incorrect and now I don't feel like making new one because the question is already answered.
How can I solve or approach this problem?
geometry euclidean-geometry triangle
asked Dec 19 '18 at 16:13
PeroPero
1207
$endgroup$
add a comment |
$begingroup$
Given right angled triangle $triangle ABC$ with right angle at point C.
Let points $D, E$ lie on $AB$, such that $|BC|=|BD|$ and $|AC|=|AE|$.
Let point $F$ be orthogonal projection of point $D$ onto $AC$ and let point $G$ be orthogonal projection of point $E$ onto $BC$.
Prove $|DE|=|DF|+|EG|$.
What I've got so far:
EDIT: Had a picture here, but I removed it because it was incorrect and now I don't feel like making new one because the question is already answered.
How can I solve or approach this problem?
geometry euclidean-geometry triangle
asked Dec 19 '18 at 16:13
PeroPero
1207
$endgroup$
2
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22
add a comment |
$begingroup$
Given right angled triangle $triangle ABC$ with right angle at point C.
Let points $D, E$ lie on $AB$, such that $|BC|=|BD|$ and $|AC|=|AE|$.
Let point $F$ be orthogonal projection of point $D$ onto $AC$ and let point $G$ be orthogonal projection of point $E$ onto $BC$.
Prove $|DE|=|DF|+|EG|$.
What I've got so far:
EDIT: Had a picture here, but I removed it because it was incorrect and now I don't feel like making new one because the question is already answered.
How can I solve or approach this problem?
geometry euclidean-geometry triangle
asked Dec 19 '18 at 16:13
PeroPero
1207
$endgroup$
Given right angled triangle $triangle ABC$ with right angle at point C.
Let points $D, E$ lie on $AB$, such that $|BC|=|BD|$ and $|AC|=|AE|$.
Let point $F$ be orthogonal projection of point $D$ onto $AC$ and let point $G$ be orthogonal projection of point $E$ onto $BC$.
Prove $|DE|=|DF|+|EG|$.
What I've got so far:
EDIT: Had a picture here, but I removed it because it was incorrect and now I don't feel like making new one because the question is already answered.
How can I solve or approach this problem?
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
asked Dec 19 '18 at 16:13
PeroPero
1207
asked Dec 19 '18 at 16:13
PeroPero
1207
edited Dec 19 '18 at 17:49
Pero
asked Dec 19 '18 at 16:13
PeroPero
1207
asked Dec 19 '18 at 16:13
PeroPero
1207
asked Dec 19 '18 at 16:13
PeroPero
1207
1207
2
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22
add a comment |
2
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22
2
2
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $angle ABC = 2x$, then $$ angle BDC = angle DCB = 90-x$$ and so $angle DCX = x$. Clearly we have $$angle ACX = 90-angle XCB = 2x,$$ so $angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
edited Dec 19 '18 at 17:49
Pero
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
$endgroup$
add a comment |
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$begingroup$
Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $angle ABC = 2x$, then $$ angle BDC = angle DCB = 90-x$$ and so $angle DCX = x$. Clearly we have $$angle ACX = 90-angle XCB = 2x,$$ so $angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
edited Dec 19 '18 at 17:49
Pero
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
$endgroup$
add a comment |
$begingroup$
Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $angle ABC = 2x$, then $$ angle BDC = angle DCB = 90-x$$ and so $angle DCX = x$. Clearly we have $$angle ACX = 90-angle XCB = 2x,$$ so $angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
edited Dec 19 '18 at 17:49
Pero
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
$endgroup$
add a comment |
$begingroup$
Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $angle ABC = 2x$, then $$ angle BDC = angle DCB = 90-x$$ and so $angle DCX = x$. Clearly we have $$angle ACX = 90-angle XCB = 2x,$$ so $angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
edited Dec 19 '18 at 17:49
Pero
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
$endgroup$
Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $angle ABC = 2x$, then $$ angle BDC = angle DCB = 90-x$$ and so $angle DCX = x$. Clearly we have $$angle ACX = 90-angle XCB = 2x,$$ so $angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
edited Dec 19 '18 at 17:49
Pero
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
edited Dec 19 '18 at 17:49
Pero
1207
edited Dec 19 '18 at 17:49
Pero
1207
edited Dec 19 '18 at 17:49
Pero
1207
1207
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
answered Dec 19 '18 at 17:03
greedoidgreedoid
40.1k114799
40.1k114799
add a comment |
add a comment |
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2
$begingroup$
Just checking: Which national competition is this? (If it's on-going, this question should be closed until the contest concludes.)
$endgroup$
– Blue
Dec 19 '18 at 16:20
$begingroup$
It's from 17.3.2018 from Macedonia.
$endgroup$
– Pero
Dec 19 '18 at 16:22