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

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 $$CD$$.

(Written before the problem statement was corrected)

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

In this case $$AD>BC$$ but $$AK.

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:

$$AK>KB$$

• @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. – Oldboy Dec 16 '18 at 12:35
• My is dowvoted also – Aqua Dec 16 '18 at 12:38
• 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. – thomas21 Dec 16 '18 at 13:52
• @thomas21 I have edited my answer. – Oldboy Dec 16 '18 at 14:09
• Quite strange how they deleted all comments on my problem post @Oldboy – thomas21 Dec 16 '18 at 16:37

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: $$K = (0,{2kac\over b-a})$$

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:

• I apologize, I expected to see the key remark somewhere at the top. – Oldboy Dec 16 '18 at 10:43
• Analysis take me to that. I didn't know it at a begining. However coordinate system helped me a lot at analysis. – Aqua Dec 16 '18 at 10:45
• Once again, I am sorry for pulling my gun so quickly :) Upvoting your answer now. – Oldboy Dec 16 '18 at 10:57
• 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. – thomas21 Dec 16 '18 at 13:52
• Then this prove settles everything. – Aqua Dec 16 '18 at 13:54