Prove that $\frac{[ABC]}{[XYZ]}=\frac{2R}{r}$, where $X$, $Y$, $Z$ are the points where the incircle of $\triangle ABC$ meets the sides

areaeuclidean-geometrygeometryratiotriangles

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:

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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 $\angle A_2$

Best Answer

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.

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For $\triangle ABC$,

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

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

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

  4. The area formula: $[ABC] = \dfrac {1}{2}ab\sin C$.

Then, $[ABC] = … = 2R^2 (\sin A)(\sin B)(\sin C)$.

Also, $[XYZ] = … = 0.5r^2(\sin A + \sin B + \sin C)$

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

  2. 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 $\dfrac {[IYZ]}{[ABC]} = \dfrac {r^2}{bc} = \dfrac {ar^2}{abc}$

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