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

euclidean-geometrygeometryquadrilateral

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 $CD$.

Best Answer

(Written before the problem statement was corrected)

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

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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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