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Exercise 137
Triangle $ABC$ has $AB=13$ and $AC=20$, and the altitude to $\overline{BC}$ has length $12$. What is the sum of the two possible values of $BC$?
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Solution
  1. Let $D$ be the foot of the altitude from $A$. $BD=\sqrt{169-144}=5$, $DC=\sqrt{400-144}=16$.
  2. If $D$ lies between $B,C$: $BC=BD+DC=21$. If $\triangle ABC$ is obtuse at $B$: $BC=DC-BD=11$.
  3. Sum $=21+11=\boxed{32}$.
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