Given a trapezoid with base $AD$ larger than side $CD$. The bisector of $angle D$ meets $AB$ at $K$. Prove...












7












$begingroup$



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 $CD$.











share|cite|improve this question











$endgroup$

















    7












    $begingroup$



    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 $CD$.











    share|cite|improve this question











    $endgroup$















      7












      7








      7


      4



      $begingroup$



      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 $CD$.











      share|cite|improve this question











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 12:57







      thomas21

















      asked Dec 16 '18 at 9:28









      thomas21thomas21

      159112




      159112






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          (Written before the problem statement was corrected)



          $AK>KB$ is simply not true in all cases:



          enter image description here



          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:



          $$AK>KB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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



















          5












          $begingroup$

          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 $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$



          enter image description here



          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          (Written before the problem statement was corrected)



          $AK>KB$ is simply not true in all cases:



          enter image description here



          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:



          $$AK>KB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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
















          6












          $begingroup$

          (Written before the problem statement was corrected)



          $AK>KB$ is simply not true in all cases:



          enter image description here



          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:



          $$AK>KB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @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














          6












          6








          6





          $begingroup$

          (Written before the problem statement was corrected)



          $AK>KB$ is simply not true in all cases:



          enter image description here



          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:



          $$AK>KB$$






          share|cite|improve this answer











          $endgroup$



          (Written before the problem statement was corrected)



          $AK>KB$ is simply not true in all cases:



          enter image description here



          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:



          $$AK>KB$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 14:08

























          answered Dec 16 '18 at 10:33









          OldboyOldboy

          8,1851936




          8,1851936












          • $begingroup$
            @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










          • $begingroup$
            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











          5












          $begingroup$

          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 $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$



          enter image description here



          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          5












          $begingroup$

          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 $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$



          enter image description here



          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          5












          5








          5





          $begingroup$

          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 $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$



          enter image description here



          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$



          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 $c$: $$A=(a,ka);;;;;;C= (c,-ck)$$



          enter image description here



          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:



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 11:01









          Sik Feng Cheong

          1579




          1579










          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


















          • $begingroup$
            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


















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