### Instruction: In isosceles triangle $\triangle ABC$ we have $AB=AC=4$.  The altitude from $B$ meets $\overline{AC}$ at $H$.  If $AH=3(HC)$ then determine $BC$.### Response: I notice that the problem involves an isosceles triangle and an altitude, which are usually helpful for finding similar triangles or right triangles.
I draw a sketch of the triangle and label the lengths I know: $AB=AC=4$ and $AH=3(HC)$.
I also mark the right angle at $H$ since the altitude is perpendicular to the base.
I see that $\triangle ABC$ is divided into two right triangles, $\triangle ABH$ and $\triangle BCH$.
I wonder if I can use the Pythagorean theorem or some similarity ratio to find $BC$.
I try to find $AH$ and $HC$ in terms of $AC$.
I use the given ratio $AH=3(HC)$ to write $AC=AH+HC=3(HC)+HC=4(HC)$.
I divide both sides by $4$ to get $HC=\frac{AC}{4}=\frac{4}{4}=1$.
Then I plug in $HC=1$ into $AH=3(HC)$ to get $AH=3(1)=3$.
I check that $AH+HC=AC$ as expected.
I use the Pythagorean theorem on $\triangle ABH$ to write $AB^2=AH^2+BH^2$.
I plug in the values I know: $4^2=3^2+BH^2$.
I solve for $BH$ by subtracting $9$ from both sides and taking the square root: $BH=\sqrt{4^2-9}=\sqrt{7}$.
I notice that $BC$ is the hypotenuse of $\triangle BCH$, so I use the Pythagorean theorem again to write $BC^2=BH^2+HC^2$.
I plug in the values I know: $BC^2=(\sqrt{7})^2+1^2$.
I simplify and solve for $BC$ by taking the square root: $BC=\sqrt{7+1}=\sqrt{8}$.