# 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 $$\angle A_2$$

• What have you tried? Mar 26 '20 at 16:05
• I know that $[ABC]=\frac{abc}{4R}$ and $[XYZ]=\frac{xyz}{4r}$ but what's next it looks even more complicated than original one Mar 27 '20 at 16:28
• Theorem 36 is true not just when $\angle A_1=\angle A_2$, it is also true when $\angle A_1+\angle A_2=180^\circ$. This is the situation here. For example, $\angle YIZ+\angle BAC=180^\circ$. Mick's answer below has a proof of this. Did you read his answer? Apr 7 '20 at 9:20

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 $$\triangle ABC$$,

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

• "In this site, I am not supposed to solve the problem completely for you." Is it a general rule or concerns only you (and possibly OP author)?
– user
Mar 29 '20 at 14:39
• @user I don't know about your or Mick's takes on that, but there are users who will downvote answers simply because the question doesn't show enough effort. And they downvote the answers regardless of how useful, correct, or well written the answers are. Mar 29 '20 at 16:18
• @user It is not a rule in general but is a common understanding for not just a few. As far as I can remember, I saw comments saying that they don't want this to be a homework doing site.
– Mick
Mar 29 '20 at 16:30
• @Mick Thanks for your effort.But I want to you to check over the proof given in my textbook and I don't quite understand it.Check my question for edit Mar 30 '20 at 12:21
• @MathematicalCuriosity I have added a picture in the middle and a further explanation at the bottom. The 'equal angles' come from the fact (1) and fact (2) that I have mention in the main text. You are welcome to raise more questions if you have other doubts.
– Mick
Mar 30 '20 at 13:18