Fermat point

$ \displaystyle \Delta ABC$ is a $ \displaystyle {{30}^{\circ }}-{{60}^{\circ }}$ right triangle with $ \displaystyle \angle A={{30}^{\circ }}\ ,\ \angle B={{60}^{\circ }}$. If P is a Fermat point of $ \displaystyle \Delta ABC$ , prove that AP , BP , CP are in a G.P.

Solution

$ \displaystyle \text{In}\ \Delta FCB\ \text{and}\ \Delta ACD\ ,$ 

$ \displaystyle \ \ \ \ \ FC\ =\ AC$ 

$ \displaystyle \ \ \ \ \ CB=CD$ 

$ \displaystyle \angle FCB=\angle ACD$ 

$ \displaystyle \Delta FCB\cong \Delta ACD\ \ (S.A.S)$ 

$\displaystyle \angle FBC=\angle ADC=\beta $ 

$ \displaystyle \therefore \ CDBP\ \text{is}\ \text{cyclic}\ .$

$ \displaystyle \angle PCB=\angle PDB=\theta $ 

$ \displaystyle AB\parallel \ CD\ \ (\because \ \angle BCD=\angle ABC)$

$ \displaystyle \therefore \ \angle ADC=\angle BAD\ \ (\because \text{alternate}\ \angle \text{s})$ 

$ \displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$ 

$ \displaystyle \ \ \ \ \ \frac{{BP}}{{\sin \beta }}=\frac{{AP}}{{\sin \theta }}$ 

$ \displaystyle \ \ \ \ \ \ \ \frac{{BP}}{{AP}}=\frac{{\sin \beta }}{{\sin \theta }}\ \ \ \ \ \ \ ........(1)$ 

$ \displaystyle \ \ \ \ \ \ \frac{{CP}}{{\sin \beta }}=\frac{{BP}}{{\sin \theta }}$

$ \displaystyle \ \ \ \ \ \ \ \frac{{CP}}{{BP}}=\frac{{\sin \beta }}{{\sin \theta }}\ \ \ \ \ \ \ ........(2)$ 

$ \displaystyle \text{From}\,(1)\ \ \text{and}\ (2)\ ,\ \ \frac{{BP}}{{AP}}=\ \frac{{CP}}{{BP}}$ 

$ \displaystyle \therefore \ AP,BP,CP\ \text{are}\ \text{in}\ \text{a}\ G.P.\ $