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The triangle $DEF$ circumscribes the three escribed circles of triangle $ABC$.

Prove that

$$\frac{EF}{a\,\cos A} = \frac{FD}{b\,\cos B} = \frac{DE}{c\,\cos C}$$

Figure

MvG
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vamsi
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2 Answers2

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Significant Edit to simplify the argument.

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In the diagram, we note that, for instance, $\overleftrightarrow{PQ}$ bisects the angle formed by $\overleftrightarrow{AB}$ and $\overleftrightarrow{DE}$. In fact, $\overleftrightarrow{PQ}$ is a line of symmetry through elements bounded by $\overleftrightarrow{AB}$ and $\overleftrightarrow{DE}$; in particular, this line of symmetry separates $\triangle ABC$ from a mirror image, $\triangle A_F B_F C$. Likewise, $\overleftrightarrow{QR}$ and $\overleftrightarrow{RP}$ give us mirror images $\triangle AB_DC_D$ and $\triangle A_EBC_E$.

As a result, $\triangle DA_EA_F$ is an isosceles triangle with base angles of size $A$. This, along with corresponding observations about $\triangle EB_FB_D$ and $\triangle FC_DC_E$, tells us $$\angle D = 180° - 2\;\angle A \qquad \angle E = 180° - 2\;\angle B \qquad \angle F = 180° - 2\;\angle C $$

Now, turning our attention to $\triangle DEF$, we invoke the Law of Sines: $$\frac{|\overline{EF}|}{\sin D} = \frac{|\overline{FD}|}{\sin E} = \frac{|\overline{DE}|}{\sin F}$$ Therefore, as $\sin(180°-\theta) = \sin\theta$, $$\frac{|\overline{EF}|}{\sin 2A} = \frac{|\overline{FD}|}{\sin 2B} = \frac{|\overline{DE}|}{\sin 2C}$$ and, as $\sin 2\theta = 2\sin\theta\cos\theta$, $$\frac{|\overline{EF}|}{2\sin A \cos A} = \frac{|\overline{FD}|}{2\sin B \cos B} = \frac{|\overline{DE}|}{2\sin C\cos C} \tag{$\star$}$$

Recall that the Law of Sines for $\triangle ABC$ —say, with circumdiameter $m$— implies $$a = m \sin A \qquad b = m \sin B \qquad c = m \sin C$$ This allow us to write $(\star)$ as $$\frac{m\;|\overline{EF}|}{2a\;\cos A} = \frac{m\;|\overline{FD}|}{2b\;\cos B} = \frac{m\;|\overline{DE}|}{2c\;\cos C}$$

Dividing-through by $m/2$ gives the result. $\square$

Blue
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  • Very nice! I'm glad you managed to write your answer before I had a chance to try squeezing my own coordinate-based proof into the width of these MSE pages. Yours is far superior. – MvG Oct 30 '14 at 08:20
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    @MvG: Thanks. My scratch work had a ridiculous amount of convoluted trigonometry in it (and my diagram had a lot more angles marked), before I realized what was "really" going on. :) I tried coordinates, too; not fun. For a while, I thought the key was to observe that each of the three triangles sharing a "vertical angle" with $\triangle ABC$ is in fact congruent to $\triangle ABC$. (Come to think of it, the angle chase is easier using this fact. I need to make a new edit. :) – Blue Oct 30 '14 at 18:23
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    My approach was starting with three rational parametrizations of the unit circle, which I take as the touching points for the incircle of $ABC$. Then everything else follows from that, and can be expressed in terms of rational functions. I was surprised to find that the final expression could even avoid square roots. It was invariant under permutations of the three variables, very much like this answer of mine. The final rational expression factors nicely, but along the way there are some huge polynomials; no fun without a CAS. – MvG Oct 30 '14 at 19:00
  • @MvG: How did we survive before CASes? :) – Blue Oct 30 '14 at 19:02
  • Nice answer Blue, to an interesting question. An idle query: So if we keep making bigger triangles $HIJ$ expanding in the same way with direct tangents, will this ratio remain the same, and ad infinitum? – Narasimham Oct 19 '15 at 10:00
  • @Blue , I don't understand how PQ is a line of symmetry and how those 2 triangles are congruent?can you please explain in detail – Vamsi Spidy Oct 19 '15 at 14:39
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I think just because $PQ $ is not extended backwards you are not seeing a situation of symmetry between direct and inverse tangents. I have relabeled for convenience, extended the direct tangents and line of centers to be concurrent at $F$ enabling comparison of angles.

Direct& Inverse Tgts

Narasimham
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  • Excellent, you made the triangles similar, but to tell that they are congruent you need to prove at least 1 pair of sides are equal. It can be proved OV and Ov are equal because OVF and OvF are congruent by ASA or AAS criteria. Now since one pair of sides are equal and corresponding angles are equal, you can say that they are congruent triangles – Vamsi Spidy Oct 20 '15 at 02:44
  • Right,normally teachers leave out such end details to students to be filled in as exercises. :) – Narasimham Oct 21 '15 at 01:48