In triangle ABC,we have $AB>AC$ . If A' is the mid point of BC,AD is the altitude through A and if the internal and external bisectors of angle A meet at BC at X and X' respectively ,prove that $A'D = (c^2-b^2)/2a$
I have tried to use obtuse angle theorem to find height then solve the question but stuck on double variable.
Sir here I tried to find height of triangle ABC but failed. Please help!
Since $c>b$, we obtain $\gamma>\beta,$ which says that $\angle ABC$ is an acute angle.
Thus, $$BD=c\cos\beta=\frac{c(a^2+c^2-b^2)}{2ac}=\frac{a^2+c^2-b^2}{2a}.$$ Also, me see that $BD>BA'$ because $c>b$.
Thus, $$A'D=BD-BA'=\frac{a^2+c^2-b^2}{2a}-\frac{a}{2}=\frac{c^2-b^2}{2a}.$$