Triangle $ ABC $, has $ \angle ACB = 60^{\circ} $. The point $ E $ lies on $ AB $ such that $ CE $ bisects $ \angle ACB $. The point $ D $ lies on $ BC $ such that $ AD \perp BC $. The point $ F $ lies on $ AC $ such that lines $ AD $, $ CE $ and $ BF $ are concurrent at a point $ O $. Lines $ EF $ and $ AD $ intersect at $ G $. Line $ BG $ is extended and a point $ K $ lies on it such that $ AK \perp BK $.
How can we prove that $ |BK|^2 = |BD|^2 + |DK|^2 + 2 |AK|^2 $?
There is a solution using coordinate geometry to this here. It uses coordinate geometry to establish that $ |AG| = |DG| $ i.e. $ G $ is the midpoint of $ AD $, and from there, 'normal' geometry techniques (trig, angle theorems etc) can be used to finish it off.
My question is, can we find a proof of this using only 'normal' geometry?
My thoughts: I spent a while on this and couldn't do it. I wondered if maybe this problem is linked to the famously hard Langley's adventitious angles type of problem, usually requiring a difficult-to-spot construction. I can spot two potentially 'adventitious quadrangles' in this problem – $ BEFC $ and $ AKBD $. However, these kinds of problems can also be solved with the trigonometric Ceva's theorem, which I tried with no success on this occasion.
I'm also familiar with things like the angle bisector/midpoint/Appolonius'/Stewart's/Menelaus'/Ceva's two theorems, but couldn't make it work with any of them. In addition to the sine/cosine rules, a solution using these techniques is what I'm looking for. Thanks.
Best Answer
By angle bisector theorem we have $\dfrac{AO}{OD}=\dfrac{AC}{CD}=2$.
Now, we can exploit a well-known fact in this configuration the quadruple $A,O,G,D$ is harmonic. This means that $\dfrac{AG}{GO}=\dfrac{AD}{DO}$ which together with $\dfrac{AO}{OD}=2$ implies that $G$ is the midpoint of $AD$.
If you're not familiar with harmonic quadruples you can prove that $\dfrac{AG}{GO}=\dfrac{AD}{DO}$ directly using Menelaos and Ceva. Menelaos for triangle $AOC$ and transversal $GEF$ gives $\dfrac{AG}{GO}\cdot \dfrac{OE}{EC}\cdot\dfrac{CF}{FA}=1$. Ceva for triangle $AOC$ and cevians $AE$, $OF$ and $CD$ gives $\dfrac{AD}{DO}\cdot\dfrac{OE}{EC}\cdot\dfrac{CF}{FA}=1$. This shows that $\dfrac{AG}{GO}=\dfrac{AD}{DO}$.
This shows that $G$ is the midpoint of $AD$.