euclidean-geometrygeometrytriangles
Let $AD$ be the angle bisector of angle $A$ in $\Delta ABC$. Prove that $$BD = BC \cdot \frac{AB}{AB + AC}$$
Hello,
I was doing some geometry and got stuck in this question. I tried using the angle bisector theorem and I know it will be used somewhere in this problem but can’t really get it right. Can you please help me with this question? I would be grateful if you did.
Thanks.
Compare the distances of $O$ to the three sides of the original triangles.
Having $O$ on an angle bisector means exactly that some of these distances are the same. This may be a theorem that you already know or you have to prove it using congruence with the help of the equal half-angles.
Therefore, you can conclude $dist(O,BC)=dist(O,AB)$ and $dist(O,CA)=dist(O,CB)$ which gives $dist(O,AB) = dist(O,AC)$.
So $O$ also lies on an angle bisector of $A$ and since it lies inside the angle at $A$ of the triangle, this has to be the internal bisector.
It is easier to prove the following equivalent version:
Let $N$ be the point of intersection between the circumcircle of $ABC$ (centered at $O$) and the angle bisector from $A$. Then $N$ lies on the perpendicular bisector of $BC$.
Proof: since $\widehat{BAN}=\widehat{NAC}$, the $BN$ and $NC$ arcs in the circumcircle of $ABC$ have the same length, since they are subtended by equal angles at $O$. It follows that the segments $BN$ and $NC$ have the same length, and since $OB=OC=R$, the line through $O$ and $N$ is the perpendicular bisector of $BC$.
Best Answer
Using the theorem about the internal bisector, we get $$\frac{BD}{DC}=\frac{c}{b}$$ so we get $$BD=DC\cdot \frac{c}{b}$$ Using that $$DC=a-BD$$ We get $$BD=(a-BD)\cdot \frac{c}{b}$$ and we obtain $$BD(1+\frac{c}{b})=\frac{ac}{b}$$ Can you finish?