Neculai STANCIU (Buzău) in the MATHEMATICAL JOURNAL // Neculai STANCIU (Buzău) în GAZETA MATEMATICĂ

 In the recent issue of the Exercise Supplement (aka SGM) he proposed the Problem S:E25.199. The problem statement is :

                    "Consider  $\Delta ABC$  a triangle such that  $\measuredangle A=2\measuredangle C$.

                      Prove that  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

Neculai STANCIU, Buzău"

Solution CiP

We construct  $AD$ - the bisector of angle  $\widehat{BAC}$. We have 

$$\measuredangle CAD=\measuredangle BAD=\frac{\measuredangle BAC}{2}=\measuredangle C$$

          We now construct  $BE\parallel AD,\;E\in AC$. We have for the angles formed with the secant $AB$  that

$\measuredangle BAD=\measuredangle ABE=\measuredangle C \tag{1}$

and the exterior angle  $\widehat{BAC}$ of triangle  $\Delta ABE$ shows us that

$\measuredangle BAC=\measuredangle ABE+\measuredangle AEB\Leftrightarrow 2\measuredangle C\overset{(1)}{=}\measuredangle C+\measuredangle AEB\Rightarrow \measuredangle AEB=\measuredangle C$

From all the angles equal to  $C$  in the figure, we see that we have the isosceles triangles  $BAE$  and  $CBE$, so

$ AB=AE\;,\;BC=BE\tag{2}$

Constructing the height  $AF$  in the isosceles triangle  $ABE,\;AF\perp BE$, it will also be the median, so  $BF=EF=BE/2$. Then, in triangle  $ABF$

$\cos C\underset{(1)}{=}\cos \widehat{ABF}=\frac{BF}{AB}=\frac{BE/2}{AB}\underset{(2)}{=}\frac{BC/2}{AB}=\frac{BC}{2\cdot AB}$,  which is equivalent to the statement  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

$\blacksquare$

          Remark CiP  The solution does not use any advanced trigonometry formulas, only definitions, otherwise from the Law of Sines in triangle  $ABC$  we could easily obtain

$\frac{AB}{\sin C}=\frac{BC}{\sin A}\Rightarrow \frac{AB}{BC}=\frac{\sin C}{\sin A}=\frac{\sin C}{\sin 2C}=\frac{\sin C}{2\sin C \cdot \cos C}=\frac{1}{2\cos C}$.