Hello thereddevils Originally Posted by

**thereddevils** THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad

Prove that triangle ABC and BPC are similar triangles . Hence deduce that

(1) BC^2=CP.CA

(2) $\displaystyle \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$

I am only stucked with (2) , i am ok with the rest . THanks .

(2) Suppose $\displaystyle BP=BC=AP = d$. Then in $\displaystyle \triangle BPC$$\displaystyle \angle BCP = \tfrac{2\pi}{5}$

$\displaystyle \Rightarrow PC = 2d\cos(\angle BCP) = 2d\cos(\tfrac{2\pi}{5})$

So, using the result in (1):$\displaystyle d^2 = 2d\cos(\tfrac{2\pi}{5})\Big(d+2d\cos(\tfrac{2\pi}{ 5})\Big)$

$\displaystyle \Rightarrow 1 = 2\cos(\tfrac{2\pi}{5})+4\cos^2(\tfrac{2\pi}{5})$

Solving the quadratic, taking the positive root gives$\displaystyle \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}$

$\displaystyle \Rightarrow 2\cos^2(\tfrac{\pi}{5})-1 = \frac{\sqrt5-1}{4}$

$\displaystyle \Rightarrow \cos^2(\tfrac{\pi}{5})= \frac{\sqrt5+3}{8}$ $\displaystyle =\frac{5+2\sqrt5+1}{16}$

$\displaystyle =\frac{(\sqrt5+1)^2}{16}$

$\displaystyle \Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}$

Grandad