1. ## 2 triangles

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

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

2. if your diagram is right, then BC = BP is wrong.

3. Originally Posted by ukorov
if your diagram is right, then BC = BP is wrong.
why is it so

4. Originally Posted by thereddevils
why is it so
oh alright not wrong actually, but could have been more detailed and more accurate.

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

I am only stucked with (2) , i am ok with the rest . THanks .
(2) Suppose $BP=BC=AP = d$. Then in $\triangle BPC$
$\angle BCP = \tfrac{2\pi}{5}$

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

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

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

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

$=\frac{(\sqrt5+1)^2}{16}$
$\Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}$

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

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

7. Hello ukorov
Originally Posted by ukorov
why $\cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1$ ??
Because $\cos2\theta = 2\cos^2\theta -1$.