# Cevians And Semicircles

### Problem

In $\Delta ABC$ $AA',$ $BB',$ and $CC'$ are concurrent cevians. Form semicircles on $AC',$ $BC',$ and so on, externally to $\Delta ABC.$ Draw common external tangents to each pair of the semicircles on the same side of the triangle. Consider inscribed angles defined by the diameters and the chords from the vertices of the triangle to the points of tangency, as shown:

Prove that

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

### Hint

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

### Solution

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

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

### Acknowledgment

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

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