Today I've encountered a question like this;
In a triangle $BFC$ angles $<B=20^\circ,\quad <F=110^\circ,\quad <C=50^\circ$,
There are points $D$(on BC) $\text{and}$ $K$(on BF) such that $|BD|=|BK|=a+b$ and $|DC|=a$, assuming $a>b$. How do we show that $|KF|=b$?,
This question is more abstract than the questions I generally deal with , although I've given it a try and my efforts are;
Creating a point $G$ such that points $D,K,G$ are linear (1)
Drawing a line from $B$ to $G$ and notice that $<GKB=30^\circ$ (2)
Seeing that $|GD|=a+b$ too (3)
Having discovered $<BGK=50^\circ$ drawing a line from $G$to $F$ (4)
Trying to find $<KGF$ then expect to use the cosine theorem or perhaps another relation (This is where I am stuck),
I've begun this attempt of a solution with expectations like drawing the replica's or symmetric's of several parts in the triangle or finding a trigonometric relation, but I failed.
What do you suggest?
The property is false.
A counterexample.-If $(a,b)=(2,1)$ then $\overline{BD}=\overline{BK}=3$ and $\overline{BC}=3+2=5$. Therefore if $\overline{KF}=b=1$ then $\overline{BF}=4$.
But then (keeping in mind that $\sin 110^{\circ}=\sin 20^{\circ}$)
$$\frac{4}{\sin 50^{\circ}}=\frac{5}{\sin 20^{\circ}}$$ This is not true.