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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 $CD$.

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(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 :

$$\frac{AK}{KB}=\frac{AD}{EB}=\frac{AD}{CD-BC}>1\tag{1}$$ 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:

$$AK>KB$$

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  • $\begingroup$ @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 $\endgroup$ – Aqua 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
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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: $$ K = (0,{2kac\over b-a})$$

Now we can calculate \begin{eqnarray}AK^2-KB^2 &=& a^2+k^2a^2{(a-b+2c)^2\over (b-a)^2}-b^2-k^2b^2{(a-b+2c)^2\over (b-a)^2}\\ &= &(a^2-b^2)(1+k^2{(a-b+2c)^2\over (b-a)^2})\\ &\geq &0 \end{eqnarray} So $AK> BK$ ONLY IF $a>|b|$. If $|b|>a$ then we have:

enter image description here

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  • $\begingroup$ 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. $\endgroup$ – Aqua 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. $\endgroup$ – Aqua Dec 16 '18 at 13:54

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