If the bisector of $\angle A$ in triangle $ABC$ meets $BC$ at $U$, prove that $AU^2=bc(1-a^2/(b+c)^2)$. (Where a, b, c are sides opposite to $A, B, C$ respectively).
I am not able to do anything. A hint to solve this in pre college level will be appreciated.
Use angle bisector theorem, BU = kc and CU = kb for some non-zero constant k.
Apply consine law to $\triangle BAU$ and $\triangle CAU$.
Try to eliminate the term involving $ \cos \dfrac {A}{2}$.
Note that $k = ... = \dfrac {a}{b + c}$.
From (2.) $(kb)^2 = AU^2 + b^2 - 2AUb \cos \dfrac {A}{2}$
Multiply throughout by c to get $c(kb)^2 = cAU^2 + cb^2 - 2AUbc \cos \dfrac {A}{2}$.
Do the similar trick to the other equation to get $b(kc)^2 = bAU^2 + bc^2 - 2AUbc \cos \dfrac {A}{2}$.
Do (3.) by subtraction.