Prove that $angle DAP=angle CAB$ in a parallelogram $ABCD$
$begingroup$
Let $ABCD$ be a parallelogram, and let $K$ be on $BC$ and $L$ on $CD$ so that $BKcdot BC=DLcdot DC$. Let point $P$ be where $DK$ and $BL$ intersect. Prove that $angle DAP=angle CAB$ (angles $DAP$ and $CAB$ are equal).
I got that $ADL$ and $ABC$ are similar so it is enough to prove that $angle LAP=angle CAK$, but I think there is an better way to use the given statement.
geometry euclidean-geometry projective-geometry geometric-transformation
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
$endgroup$
|
show 5 more comments
$begingroup$
Let $ABCD$ be a parallelogram, and let $K$ be on $BC$ and $L$ on $CD$ so that $BKcdot BC=DLcdot DC$. Let point $P$ be where $DK$ and $BL$ intersect. Prove that $angle DAP=angle CAB$ (angles $DAP$ and $CAB$ are equal).
I got that $ADL$ and $ABC$ are similar so it is enough to prove that $angle LAP=angle CAK$, but I think there is an better way to use the given statement.
geometry euclidean-geometry projective-geometry geometric-transformation
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
$endgroup$
$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
$endgroup$
– HeatTheIce
Jan 23 '15 at 15:03
$begingroup$
see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
$endgroup$
– abel
Jan 23 '15 at 16:53
$begingroup$
i have verified that the result is certainly true for a rectangle $ABCD$
$endgroup$
– abel
Jan 23 '15 at 17:19
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
$endgroup$
– Blue
Jan 23 '15 at 18:07
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
$endgroup$
– Pp..
Jan 23 '15 at 18:46
|
show 5 more comments
$begingroup$
Let $ABCD$ be a parallelogram, and let $K$ be on $BC$ and $L$ on $CD$ so that $BKcdot BC=DLcdot DC$. Let point $P$ be where $DK$ and $BL$ intersect. Prove that $angle DAP=angle CAB$ (angles $DAP$ and $CAB$ are equal).
I got that $ADL$ and $ABC$ are similar so it is enough to prove that $angle LAP=angle CAK$, but I think there is an better way to use the given statement.
geometry euclidean-geometry projective-geometry geometric-transformation
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
$endgroup$
Let $ABCD$ be a parallelogram, and let $K$ be on $BC$ and $L$ on $CD$ so that $BKcdot BC=DLcdot DC$. Let point $P$ be where $DK$ and $BL$ intersect. Prove that $angle DAP=angle CAB$ (angles $DAP$ and $CAB$ are equal).
I got that $ADL$ and $ABC$ are similar so it is enough to prove that $angle LAP=angle CAK$, but I think there is an better way to use the given statement.
geometry euclidean-geometry projective-geometry geometric-transformation
geometry euclidean-geometry projective-geometry geometric-transformation
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
edited Dec 29 '18 at 15:11
Maria Mazur
46.9k1260120
46.9k1260120
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
asked Jan 23 '15 at 14:23
HeatTheIceHeatTheIce
1,775618
1,775618
$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
$endgroup$
– HeatTheIce
Jan 23 '15 at 15:03
$begingroup$
see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
$endgroup$
– abel
Jan 23 '15 at 16:53
$begingroup$
i have verified that the result is certainly true for a rectangle $ABCD$
$endgroup$
– abel
Jan 23 '15 at 17:19
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
$endgroup$
– Blue
Jan 23 '15 at 18:07
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
$endgroup$
– Pp..
Jan 23 '15 at 18:46
|
show 5 more comments
$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
$endgroup$
– HeatTheIce
Jan 23 '15 at 15:03
$begingroup$
see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
$endgroup$
– abel
Jan 23 '15 at 16:53
$begingroup$
i have verified that the result is certainly true for a rectangle $ABCD$
$endgroup$
– abel
Jan 23 '15 at 17:19
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
$endgroup$
– Blue
Jan 23 '15 at 18:07
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
$endgroup$
– Pp..
Jan 23 '15 at 18:46
$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
$endgroup$
– HeatTheIce
Jan 23 '15 at 15:03
$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
$endgroup$
– HeatTheIce
Jan 23 '15 at 15:03
$begingroup$
see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
$endgroup$
– abel
Jan 23 '15 at 16:53
$begingroup$
see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
$endgroup$
– abel
Jan 23 '15 at 16:53
$begingroup$
i have verified that the result is certainly true for a rectangle $ABCD$
$endgroup$
– abel
Jan 23 '15 at 17:19
$begingroup$
i have verified that the result is certainly true for a rectangle $ABCD$
$endgroup$
– abel
Jan 23 '15 at 17:19
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
$endgroup$
– Blue
Jan 23 '15 at 18:07
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
$endgroup$
– Blue
Jan 23 '15 at 18:07
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
$endgroup$
– Pp..
Jan 23 '15 at 18:46
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
$endgroup$
– Pp..
Jan 23 '15 at 18:46
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
This is a solution using oblique coordinates.
The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.
Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,ell_1)$, $D=(ell_2,0)$ and $C=(ell_2,ell_1)$. Call $r:=frac{ell_1}{ell_2}$.
The line $AC$ is $y=rx$.
We want to prove that the line $y=frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.
Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(ell_2,0)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-0}{a-ell_2}x-ell_2frac{frac{a}{r}-0}{a-ell_2}$$
This intersects $BC=(y=ell_1)$ at $K=left(left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a},ell_1right)$, i.e. $$BK=left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}$$
The line from $B=(0,ell_1)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-ell_1}{a-0}x+ell_1$$
This line intersects $DC=(x=ell_2)$ at $L=left(ell_2,frac{a-rell_1}{ra}ell_2+ell_1right)$, i.e.
$$DL=frac{a-rell_1}{ra}ell_2+ell_1$$
Now we only need to divide
$$frac{BK}{DL}=frac{left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}}{frac{a-rell_1}{ra}ell_2+ell_1}=r=frac{ell_1}{ell_2}=frac{DC}{BC}$$
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
$endgroup$
add a comment |
$begingroup$
With knowledge of projective geometry this is pretty easy problem.
Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {bover a} x$$
so $L$ is lineary dependent on $K$, thus the transformation $$Kmapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line.
Clearly when $K=infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DEover DA} = {{b^2over a }over b} = {bover a} = {BCover BA}$$
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
This is a solution using oblique coordinates.
The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.
Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,ell_1)$, $D=(ell_2,0)$ and $C=(ell_2,ell_1)$. Call $r:=frac{ell_1}{ell_2}$.
The line $AC$ is $y=rx$.
We want to prove that the line $y=frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.
Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(ell_2,0)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-0}{a-ell_2}x-ell_2frac{frac{a}{r}-0}{a-ell_2}$$
This intersects $BC=(y=ell_1)$ at $K=left(left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a},ell_1right)$, i.e. $$BK=left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}$$
The line from $B=(0,ell_1)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-ell_1}{a-0}x+ell_1$$
This line intersects $DC=(x=ell_2)$ at $L=left(ell_2,frac{a-rell_1}{ra}ell_2+ell_1right)$, i.e.
$$DL=frac{a-rell_1}{ra}ell_2+ell_1$$
Now we only need to divide
$$frac{BK}{DL}=frac{left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}}{frac{a-rell_1}{ra}ell_2+ell_1}=r=frac{ell_1}{ell_2}=frac{DC}{BC}$$
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
$endgroup$
add a comment |
$begingroup$
This is a solution using oblique coordinates.
The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.
Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,ell_1)$, $D=(ell_2,0)$ and $C=(ell_2,ell_1)$. Call $r:=frac{ell_1}{ell_2}$.
The line $AC$ is $y=rx$.
We want to prove that the line $y=frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.
Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(ell_2,0)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-0}{a-ell_2}x-ell_2frac{frac{a}{r}-0}{a-ell_2}$$
This intersects $BC=(y=ell_1)$ at $K=left(left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a},ell_1right)$, i.e. $$BK=left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}$$
The line from $B=(0,ell_1)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-ell_1}{a-0}x+ell_1$$
This line intersects $DC=(x=ell_2)$ at $L=left(ell_2,frac{a-rell_1}{ra}ell_2+ell_1right)$, i.e.
$$DL=frac{a-rell_1}{ra}ell_2+ell_1$$
Now we only need to divide
$$frac{BK}{DL}=frac{left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}}{frac{a-rell_1}{ra}ell_2+ell_1}=r=frac{ell_1}{ell_2}=frac{DC}{BC}$$
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
$endgroup$
add a comment |
$begingroup$
This is a solution using oblique coordinates.
The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.
Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,ell_1)$, $D=(ell_2,0)$ and $C=(ell_2,ell_1)$. Call $r:=frac{ell_1}{ell_2}$.
The line $AC$ is $y=rx$.
We want to prove that the line $y=frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.
Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(ell_2,0)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-0}{a-ell_2}x-ell_2frac{frac{a}{r}-0}{a-ell_2}$$
This intersects $BC=(y=ell_1)$ at $K=left(left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a},ell_1right)$, i.e. $$BK=left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}$$
The line from $B=(0,ell_1)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-ell_1}{a-0}x+ell_1$$
This line intersects $DC=(x=ell_2)$ at $L=left(ell_2,frac{a-rell_1}{ra}ell_2+ell_1right)$, i.e.
$$DL=frac{a-rell_1}{ra}ell_2+ell_1$$
Now we only need to divide
$$frac{BK}{DL}=frac{left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}}{frac{a-rell_1}{ra}ell_2+ell_1}=r=frac{ell_1}{ell_2}=frac{DC}{BC}$$
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
$endgroup$
This is a solution using oblique coordinates.
The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.
Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,ell_1)$, $D=(ell_2,0)$ and $C=(ell_2,ell_1)$. Call $r:=frac{ell_1}{ell_2}$.
The line $AC$ is $y=rx$.
We want to prove that the line $y=frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.
Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(ell_2,0)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-0}{a-ell_2}x-ell_2frac{frac{a}{r}-0}{a-ell_2}$$
This intersects $BC=(y=ell_1)$ at $K=left(left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a},ell_1right)$, i.e. $$BK=left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}$$
The line from $B=(0,ell_1)$ to $P=(a,a/r)$ is given by
$$y=frac{frac{a}{r}-ell_1}{a-0}x+ell_1$$
This line intersects $DC=(x=ell_2)$ at $L=left(ell_2,frac{a-rell_1}{ra}ell_2+ell_1right)$, i.e.
$$DL=frac{a-rell_1}{ra}ell_2+ell_1$$
Now we only need to divide
$$frac{BK}{DL}=frac{left[ell_1+frac{ell_2a}{r(a-ell_2)}right]cdotfrac{r(a-ell_2)}{a}}{frac{a-rell_1}{ra}ell_2+ell_1}=r=frac{ell_1}{ell_2}=frac{DC}{BC}$$
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
edited Jan 25 '15 at 5:39
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
answered Jan 24 '15 at 5:02
Pp..Pp..
5,1791927
5,1791927
add a comment |
add a comment |
$begingroup$
With knowledge of projective geometry this is pretty easy problem.
Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {bover a} x$$
so $L$ is lineary dependent on $K$, thus the transformation $$Kmapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line.
Clearly when $K=infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DEover DA} = {{b^2over a }over b} = {bover a} = {BCover BA}$$
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
$endgroup$
add a comment |
$begingroup$
With knowledge of projective geometry this is pretty easy problem.
Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {bover a} x$$
so $L$ is lineary dependent on $K$, thus the transformation $$Kmapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line.
Clearly when $K=infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DEover DA} = {{b^2over a }over b} = {bover a} = {BCover BA}$$
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
$endgroup$
add a comment |
$begingroup$
With knowledge of projective geometry this is pretty easy problem.
Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {bover a} x$$
so $L$ is lineary dependent on $K$, thus the transformation $$Kmapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line.
Clearly when $K=infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DEover DA} = {{b^2over a }over b} = {bover a} = {BCover BA}$$
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
$endgroup$
With knowledge of projective geometry this is pretty easy problem.
Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {bover a} x$$
so $L$ is lineary dependent on $K$, thus the transformation $$Kmapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line.
Clearly when $K=infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DEover DA} = {{b^2over a }over b} = {bover a} = {BCover BA}$$
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
edited Dec 25 '18 at 18:46
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
answered Dec 25 '18 at 18:40
Maria MazurMaria Mazur
46.9k1260120
46.9k1260120
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$begingroup$
docs.google.com/file/d/0ByGmeK3txyClSUM5OFNoSmQ5R1E/…. Here is a picture I drew, bcs I dont have a pc nearby.
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– HeatTheIce
Jan 23 '15 at 15:03
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see if this works. extend $AP$ to meet $BC$ at $Q.$ seems that $AKQ$ is similar to $ALC$
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– abel
Jan 23 '15 at 16:53
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i have verified that the result is certainly true for a rectangle $ABCD$
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– abel
Jan 23 '15 at 17:19
$begingroup$
For those looking to construct this figure: Take $O$ a point on the perpendicular bisector of $overline{BD}$, and consider $bigcirc O$ that passes through $C$. Then $K$ and $L$ are the other points where $bigcirc O$ passes through the specified edges of the parallelogram.
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– Blue
Jan 23 '15 at 18:07
$begingroup$
Well, we can just draw a ray from A forming an angle with AC equal to CAB, pick any point P on that ray, and join P with C and with B to get K and L. By the way, @Blue can you check your construction? I think it looks very suspicious. If we put the parallel translations K' and L' of K and L on the sides AD and AB, then I see a circle through BDK'L'. Those circles on C that Jack and you constructed, I don't see them.
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– Pp..
Jan 23 '15 at 18:46