Given $\triangle ABC$, with $AB<BC$, and angle bisector $BK$, which length is larger: $AK$ or $KC$? Seeking alternate proof.

Question

The problem

The problem gives you a triangle $ABC$, where $AB<BC$. Then it draws the angle bisector $BK$, and asks you which length is larger: $AK$ or $KC$?

This is a problem from my geometry textbook "Ģeometrija 7. - 9. klasei II. Trijstūri" (It's in latvian and there are no answers in it sadly, also it sometimes has some incredibly hard problems (I found a problem (#219) that was a much harder variant of the last problem on the 42. AMO (Atklātā Matemātikas Olimpiāde) for 8th graders) ). I'm not sure whether or not this problem is already on the internet or not (hard to search for this kind of problem). I was able to solve it in a direct way, but i still want to see if it's possible to solve it using proof by contradiction. Would really like to see some other ways of proving this!

My direct proof

By playing around with many triangles, i came to the conclusion that $AK<KC$, but we still need to prove that!

First place a point $E$ on $BC$, such that $AB=BE$ (possible since $AB<BC$). Note, that $\Delta ABK\cong \Delta EBK$, since $AB=BE$, $\angle ABK=\angle KBE$, and $BK$ is a common side. From that we can conclude that $AK=KE$ and $\angle BAK=\angle BEK$. Since $\angle BEK$ and $\angle KEC$ are supplementary, we can deduce that: $$\angle KEC=180^\mathrm{o}-\angle BEK=180^\mathrm{o}-\angle BAK$$ The sum of the interior angles of a triangle is $180^\mathrm{o}$, thus: $$\angle ACB+\angle ABC=180^\mathrm{o}-\angle BAC=180^\mathrm{o}-\angle BAK=\angle KEC$$ Since the size of an angle is always positive, $\angle ACB<\angle KEC$. Then against the larger angle will be the larger side, thus: $$KE<KC$$ Remembering that $AK=KE$, we get: $$AK<KC$$ This completes my proof.

My struggles with proof by contradiction

I tried using proof by contradiction. I am able to prove that $AK\not = KC$, but i can't seem to find a way to prove that $AK\not > KC$, without using my direct proof. There's probably a way, but if it exists, it has done a great job evading me. I'm excited to see how you'll tackle this!

Answer 1

Draw $KM\perp AB$ , $KN\perp BC$ and $BH\perp AC$

Since $BK$ is angle bisector then $KM=KN$,$\triangle BMK \cong \triangle BNK$

$$\frac{\text{area}\triangle ABK}{\text{area}\triangle CBK}=\frac{\frac{1}{2}AB\times KM}{\frac{1}{2}BC\times KN}=\frac{AB}{BC}=\frac{\frac{1}{2}AK\times BH}{\frac{1}{2}CK\times BH}=\frac{AK}{CK}$$

$$1>\frac{AB}{BC}=\frac{AK}{CK}\text{(angle bisector theorem)}$$ then $$CK>AK$$

Answer 2

Do you know/can you use (at that stage in the book) that $AB:BC=AK:KC$? (Because this will give you the answer straight away.)

If you can't, you can still prove it "on the spot", as follows.

Extend $AB$ further "after" $B$ to a point $X$ such that $BX=BC$, then conclude that $\triangle BCX$ is isosceles and so $\angle BCX=\frac{1}{2}\angle ABC=\angle KBC$ so $BK\parallel CX$ and then $AB:BC=AB:BX=AK:KC$ follows from Thales'/Intercept theorem.