Question 165
In triangle $A B C$, the bisector of $\angle B$ meets the circumcircle of $\triangle A B C$ at $D$. Prove that
$$
B D^{2} \gt B A \cdot B C
$$
$$
B D^{2} \gt B A \cdot B C
$$
MathNet 2009-part-04 — 175 problems (EN–VI) Question 165
Math Olympiad Vault · MathNet 2009-part-04 — 175 problems (EN–VI)