Given a trapezoid with base $AD$ larger than side $CD$. The bisector of $angle D$ meets $AB$ at $K$. Prove...
$begingroup$
We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$.
All that I have tried was to make such drawing in GeoGebra, which obviously showed me that $AK>KB$, even if I extend $AD$ very, very long. I think the solution should go somehow through similar triangles, but I honestly have no idea how. I would really appreciate any help you provide. To mention more, I seriously don't need the entire solution. Even a little hint would be very helpful for me, since I don't really know where to start.
EDIT: key mistake was made in the previous problem: it's not $AD$ that's larger than $BC$, but $AD$ is larger than side $CD$.
geometry euclidean-geometry quadrilateral
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
$endgroup$
add a comment |
$begingroup$
We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$.
All that I have tried was to make such drawing in GeoGebra, which obviously showed me that $AK>KB$, even if I extend $AD$ very, very long. I think the solution should go somehow through similar triangles, but I honestly have no idea how. I would really appreciate any help you provide. To mention more, I seriously don't need the entire solution. Even a little hint would be very helpful for me, since I don't really know where to start.
EDIT: key mistake was made in the previous problem: it's not $AD$ that's larger than $BC$, but $AD$ is larger than side $CD$.
geometry euclidean-geometry quadrilateral
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
$endgroup$
add a comment |
$begingroup$
We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$.
All that I have tried was to make such drawing in GeoGebra, which obviously showed me that $AK>KB$, even if I extend $AD$ very, very long. I think the solution should go somehow through similar triangles, but I honestly have no idea how. I would really appreciate any help you provide. To mention more, I seriously don't need the entire solution. Even a little hint would be very helpful for me, since I don't really know where to start.
EDIT: key mistake was made in the previous problem: it's not $AD$ that's larger than $BC$, but $AD$ is larger than side $CD$.
geometry euclidean-geometry quadrilateral
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
$endgroup$
We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$.
All that I have tried was to make such drawing in GeoGebra, which obviously showed me that $AK>KB$, even if I extend $AD$ very, very long. I think the solution should go somehow through similar triangles, but I honestly have no idea how. I would really appreciate any help you provide. To mention more, I seriously don't need the entire solution. Even a little hint would be very helpful for me, since I don't really know where to start.
EDIT: key mistake was made in the previous problem: it's not $AD$ that's larger than $BC$, but $AD$ is larger than side $CD$.
geometry euclidean-geometry quadrilateral
geometry euclidean-geometry quadrilateral
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
edited Dec 16 '18 at 12:57
thomas21
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
asked Dec 16 '18 at 9:28
thomas21thomas21
159112
159112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$frac{AK}{KB}=frac{AD}{EB}$$
If $AK>KB$ then :
$$frac{AK}{KB}=frac{AD}{EB}=frac{AD}{CD-BC}>1tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$frac{AK}{KB}=frac{AD}{CD-BC}gt frac{AD}{CD}gt1$$
...or:
$$AK>KB$$
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
$endgroup$
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
add a comment |
$begingroup$
One obvious way is a way with introduction of a coordinate system. Let $D$ be the origin and $DK$ the y-axis. Say $A$ is one line $y=kx$ then $C$ is on line $y=-kx$, so we have for some positive $a$ and negative $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$
Since $B$ is on a parallel with $AD$ through $C$, which has an equation $$ y = kx -2kc$$ we have, for some $b>c$: $$B= (b,k(b-2c))$$
Now the line $AB$ has an equation $$y -ka = {k(a-b+2c)over a-b}(x-a)$$
so for $x=0$ we get $y$ coordinate of $K$ and we get:
$$ K = (0,{2kacover b-a})$$
Now we can calculate
begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2over (b-a)^2}\
&= &(a^2-b^2)(1+k^2{(a-b+2c)^2over (b-a)^2})\
&geq &0
end{eqnarray}
So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
$endgroup$
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$frac{AK}{KB}=frac{AD}{EB}$$
If $AK>KB$ then :
$$frac{AK}{KB}=frac{AD}{EB}=frac{AD}{CD-BC}>1tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$frac{AK}{KB}=frac{AD}{CD-BC}gt frac{AD}{CD}gt1$$
...or:
$$AK>KB$$
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
$endgroup$
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
add a comment |
$begingroup$
(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$frac{AK}{KB}=frac{AD}{EB}$$
If $AK>KB$ then :
$$frac{AK}{KB}=frac{AD}{EB}=frac{AD}{CD-BC}>1tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$frac{AK}{KB}=frac{AD}{CD-BC}gt frac{AD}{CD}gt1$$
...or:
$$AK>KB$$
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
$endgroup$
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
add a comment |
$begingroup$
(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$frac{AK}{KB}=frac{AD}{EB}$$
If $AK>KB$ then :
$$frac{AK}{KB}=frac{AD}{EB}=frac{AD}{CD-BC}>1tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$frac{AK}{KB}=frac{AD}{CD-BC}gt frac{AD}{CD}gt1$$
...or:
$$AK>KB$$
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
$endgroup$
(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$frac{AK}{KB}=frac{AD}{EB}$$
If $AK>KB$ then :
$$frac{AK}{KB}=frac{AD}{EB}=frac{AD}{CD-BC}>1tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$frac{AK}{KB}=frac{AD}{CD-BC}gt frac{AD}{CD}gt1$$
...or:
$$AK>KB$$
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
edited Dec 16 '18 at 14:08
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
answered Dec 16 '18 at 10:33
OldboyOldboy
8,1751936
8,1751936
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
add a comment |
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
@greedoid It's not so much better because someone has downvoted my answer (without leaving a comment, which is really sad). I just think that in simple problems like this one, people expect something from Euclid, not from Descartes.
$endgroup$
– Oldboy
Dec 16 '18 at 12:35
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
My is dowvoted also
$endgroup$
– greedoid
Dec 16 '18 at 12:38
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
@thomas21 I have edited my answer.
$endgroup$
– Oldboy
Dec 16 '18 at 14:09
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
$begingroup$
Quite strange how they deleted all comments on my problem post @Oldboy
$endgroup$
– thomas21
Dec 16 '18 at 16:37
add a comment |
$begingroup$
One obvious way is a way with introduction of a coordinate system. Let $D$ be the origin and $DK$ the y-axis. Say $A$ is one line $y=kx$ then $C$ is on line $y=-kx$, so we have for some positive $a$ and negative $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$
Since $B$ is on a parallel with $AD$ through $C$, which has an equation $$ y = kx -2kc$$ we have, for some $b>c$: $$B= (b,k(b-2c))$$
Now the line $AB$ has an equation $$y -ka = {k(a-b+2c)over a-b}(x-a)$$
so for $x=0$ we get $y$ coordinate of $K$ and we get:
$$ K = (0,{2kacover b-a})$$
Now we can calculate
begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2over (b-a)^2}\
&= &(a^2-b^2)(1+k^2{(a-b+2c)^2over (b-a)^2})\
&geq &0
end{eqnarray}
So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
$endgroup$
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
add a comment |
$begingroup$
One obvious way is a way with introduction of a coordinate system. Let $D$ be the origin and $DK$ the y-axis. Say $A$ is one line $y=kx$ then $C$ is on line $y=-kx$, so we have for some positive $a$ and negative $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$
Since $B$ is on a parallel with $AD$ through $C$, which has an equation $$ y = kx -2kc$$ we have, for some $b>c$: $$B= (b,k(b-2c))$$
Now the line $AB$ has an equation $$y -ka = {k(a-b+2c)over a-b}(x-a)$$
so for $x=0$ we get $y$ coordinate of $K$ and we get:
$$ K = (0,{2kacover b-a})$$
Now we can calculate
begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2over (b-a)^2}\
&= &(a^2-b^2)(1+k^2{(a-b+2c)^2over (b-a)^2})\
&geq &0
end{eqnarray}
So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
$endgroup$
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
add a comment |
$begingroup$
One obvious way is a way with introduction of a coordinate system. Let $D$ be the origin and $DK$ the y-axis. Say $A$ is one line $y=kx$ then $C$ is on line $y=-kx$, so we have for some positive $a$ and negative $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$
Since $B$ is on a parallel with $AD$ through $C$, which has an equation $$ y = kx -2kc$$ we have, for some $b>c$: $$B= (b,k(b-2c))$$
Now the line $AB$ has an equation $$y -ka = {k(a-b+2c)over a-b}(x-a)$$
so for $x=0$ we get $y$ coordinate of $K$ and we get:
$$ K = (0,{2kacover b-a})$$
Now we can calculate
begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2over (b-a)^2}\
&= &(a^2-b^2)(1+k^2{(a-b+2c)^2over (b-a)^2})\
&geq &0
end{eqnarray}
So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
$endgroup$
One obvious way is a way with introduction of a coordinate system. Let $D$ be the origin and $DK$ the y-axis. Say $A$ is one line $y=kx$ then $C$ is on line $y=-kx$, so we have for some positive $a$ and negative $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$
Since $B$ is on a parallel with $AD$ through $C$, which has an equation $$ y = kx -2kc$$ we have, for some $b>c$: $$B= (b,k(b-2c))$$
Now the line $AB$ has an equation $$y -ka = {k(a-b+2c)over a-b}(x-a)$$
so for $x=0$ we get $y$ coordinate of $K$ and we get:
$$ K = (0,{2kacover b-a})$$
Now we can calculate
begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2over (b-a)^2}\
&= &(a^2-b^2)(1+k^2{(a-b+2c)^2over (b-a)^2})\
&geq &0
end{eqnarray}
So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
edited Dec 16 '18 at 11:01
Sik Feng Cheong
1579
1579
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
answered Dec 16 '18 at 10:17
greedoidgreedoid
41.9k1152105
41.9k1152105
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
add a comment |
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
I apologize, I expected to see the key remark somewhere at the top.
$endgroup$
– Oldboy
Dec 16 '18 at 10:43
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis.
$endgroup$
– greedoid
Dec 16 '18 at 10:45
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now.
$endgroup$
– Oldboy
Dec 16 '18 at 10:57
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
I'm so sorry that I noticed my mistake in this problem after you solved it, and I do still appreciate the work you've put in there, but if you'd want to resolve the problem that was fixed (edit of my post) it's up there.
$endgroup$
– thomas21
Dec 16 '18 at 13:52
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
$begingroup$
Then this prove settles everything.
$endgroup$
– greedoid
Dec 16 '18 at 13:54
add a comment |
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