In any isosceles triangle $ABC$ with $AB=AC$, the altitude $AD$ bisects the base $BC$ so that $BD=DC$. As shown in $\triangle ABC$, $AB=AC=25$ and $BC=14$. Determine the length of the altitude $AD$.
By Pythagoras's Theorem:
25^2 - (14/2)^2 = AD^2
AD^2 = 576
AD = sqrt(576)
AD = 24
+194 
