In Triangle ABC , AD , BE and CF are altitudes . Angle FDE = 60 and FE=8. Find BC.

Question

In triangle $\Delta ABC$ ; $AD$ , $BE$ and $CF$ are altitudes . $\angle FDE = 60^{\circ}$ and $FE = 8 \text{ cm}$.

Find $BC$ in $\text{cm}$.

Answer 1

We need the following facts:

1. The circumcircle of $\triangle FDE$ is the nine point circle of $\triangle ABC$. Thus the circumradius of $\triangle ABC$ (call it $R$) is two times the circumradius of triangle $FDE$ (call it $R'$). In math, $$R = 2R'$$
2. $$\angle FDE = 180^{\circ}-2\angle BAC$$

Fact 2 can be observed by following the proof of part 1 at this link.

Fact 1 follows from the nine point circle theorem.

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We know $BC=2R \sin \angle BAC$. Now by sine rule in $\triangle FDE$, $FE = 2R' \sin \angle FDE$, where $R'$ is the circumradius of $\triangle FDE$.

Using the above facts, we get $\angle BAC = 60^{\circ}$ and $R = 2R' = \dfrac{FE}{\sin \angle FDE} = \dfrac{16}{\sqrt{3}}$.

Therefore $$BC = 2R \sin \angle BAC = \dfrac{32}{\sqrt{3}} \sin 60^{\circ} = 16.$$