A geometry problem by A Former Brilliant Member

By Pythagorean Theorem, we have

$AC=\sqrt{8^2+4^2}=\sqrt{80}\sqrt{16(5)}=4\sqrt{5}$

Let $BE$ be the perpendicular line.

Note that triangle $ABC$ and triangle $AEB$ are similar. So

$\dfrac{EB}{8}=\dfrac{4}{4\sqrt{5}}$

$EB=\dfrac{32}{4\sqrt{5}}$

$EB=\dfrac{8}{\sqrt{5}}$

We rationalize the denominator by multiplying both the numerator and denominator by $\sqrt{5}$ , we obtain

$EB=\dfrac{8\sqrt{5}}{5}$

tan(A) = 4/8 = 1/2. sin(A) = (BE)/8, or BE = 8 sin(A). tan(A) = sin(A)/cos(A), tan^2(A) = sin^2(A)/(1- sin^2(A)), and sin^2(A) =tan^2(A)/(1 + tan^2(A)). Then sin^2(.5) = .25/1.25/ = 1/5. Then BE = 8 sqrt(1/5) = (8/5)sqrt(5). Ed Gray