Cevians And Semicircles

The diagram above shows only one chord of one semicircle per side. Draw a few more segments that could be relevant to the construction.

Consider the pair of the semicircles on side $AC.$ Let $P$ and $Q$ be their centers and $F$ and $G$ the points of tangency.

$PF\parallel QG$ because both are perpendicular to $FG.$ Therefore, $\angle APF=\angle B'QG.$ Since both triangles $APF$ and $B'QG$ are isosceles and their apex angles are equal, the two triangle are similar, implying $\alpha=\angle PAF=\angle QB'G$ so that two right triangles $AB'F$ and $B'CG$ are similar. In addition, $\Delta FGB'$ is similar to the two. This is because $B'F\perp B'G$ and $\angle B'AF=\angle B'FG.$ (The latter since inscribed $\angle B'AF$ is subtended by the arc $B'F$ whereas $\angle B'FG$ is formed by a tangent and a chord subtending the same arc. It follows that

$\displaystyle\frac{B'C}{AB'}=\frac{B'G}{AF}.$

But, from $\Delta FGB',$ $\displaystyle B'G = B'F\cdot\tan\alpha,$ and, from $\Delta AB'F,$ $AF=B'F/\tan\alpha.$ Finally,

$\displaystyle\frac{B'C}{AB'}=\frac{B'G}{AF}=\tan^{2}\alpha.$

All that remains is to apply the same considerations to the other two pairs of the semicircles and refer to Ceva's theorem

$\displaystyle\frac{B'C}{AB'}\cdot\frac{AC'}{BC'}\cdot\frac{BA'}{CA'}=1.$

This gives

$\tan^{2}\alpha\cdot\tan^{2}\beta\cdot\tan^{2}\gamma=1,$

and, since, all three angles are acute, the three tangents are positive such that taking square root is safe and yields the required equality

$\tan\alpha\cdot\tan\beta\cdot\tan\gamma=1.$

The problem has been discovered by Emmanuel Antonio José García and posted at the CutTheKnotMath facebook page.