Prove that $\frac{[ABC]}{[XYZ]}=\frac{2R}{r}$, where $X$, $Y$, $Z$ are the points where the incircle of $\triangle ABC$ meets the sides 
Prove that $$\frac{[ABC]}{[XYZ]}=\frac{2R}{r}$$
  where $[\,\_\,]$ represents area of triangle, $X,Y,Z$ are the points of contact of incircle with sides of triangle $ABC$, $R$ is circumradius, and $r$ is inradius.

The textbook proof is shown below, along with the referenced Theorem 36.

The theorem requires triangles to have equal angle, but in the question I could not find equal angles. Maybe I am wrong?


Here is my textbook proof:

Incase you are wondering what Theorem 36 is.

Theorem 36: In two triangles $A_1B_1C_1$ and $A_2B_2C_2$ we have $\angle A_1=\angle A_2$. Then their areas are proportional to rectangles contained by the sides containing $\angle A_1$ and $\angle A_2$

 A: In this site, I am not supposed to solve the problem completely for you. I can help you to finish the task by giving you the following hints/formulas. The proofs of them can be found in the internet.

For $\triangle ABC$, 


*

*$AYIX$ is cyclic implies $\angle ABC + \angle XIY = 180^0$.

*Trigo identity: $\sin (180^0 – \theta) = \sin \theta$.

*The sine law: $a = 2R \sin A$.

*The area formula: $[ABC] = \dfrac {1}{2}ab\sin C$.
Then, $[ABC] = … = 2R^2 (\sin A)(\sin B)(\sin C)$.
Also, $[XYZ] = … = 0.5r^2(\sin A + \sin B + \sin C)$


*Double angle formula: $\sin (2\theta) = 2 \sin \theta\cos \theta$.

*Relation between $r$ and $R$: $r = 4R(\sin \dfrac {A}{2})(\sin\dfrac {B}{2})(\sin\dfrac {C}{2})$.

$[IYZ] = 0.5 (IY)(IZ) \sin \angle YIZ = 0.5 r^2 \sin \angle YIZ$
Similarly, $[ABC]$
$ = 0.5 (AB)(AC) \sin \angle BAC$
$ =  0.5 (AB)(AC) \sin (180^0 - \angle YIZ) $
$=  0.5 (AB)(AC) \sin \angle YIZ$
$ = 0.5bc \sin \angle YIZ$
Then, after cancellation, we have $\dfrac {[IYZ]}{[ABC]} = \dfrac {r^2}{bc} = \dfrac {ar^2}{abc}$
