# Triangles formed by the points of contact of the sides with the excircles and by that of the sides of the triangle with the inscribed circle.

Prove that the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides is equivalent to the triangle formed by the points of contact of the sides of the triangle with the inscribed circle.

Can it be approached using Ceva's, Menalaus' or Stewart's theorems?

• Hmm. Is this true? (I'm thinking of a skinny isosceles triangle). What do you mean by "equivalent" here? – Joffan Mar 18 '17 at 1:20
• Probably, it means having same area according to the same question on Google search. – Dayal Kumar Mar 18 '17 at 1:23
• I have a GeoGebra sketch that seems to confirm that "equivalent" means "having equal area" in this context. – Blue Mar 18 '17 at 2:26
• Please give solution? – Dayal Kumar Mar 18 '17 at 10:50

I don't see a clever way to invoke Ceva, Menelaus, or Stewart here. Nevertheless, there's a general principle at work. Consider $\triangle ABC$ with points $D$, $E$, $F$ on appropriate sides at shown ...

... where we define $$p := \frac{|\overline{BD}|}{|\overline{BC}|} \qquad q := \frac{|\overline{CE}|}{|\overline{CA}|} \qquad r := \frac{|\overline{AF}|}{|\overline{AB}|} \tag{1}$$

(These are not Ceva-Menelaus ratios.) Then, for instance, since $\triangle AEF$ shares an angle with $\triangle ABC$, but the corresponding sides enclosing that angle are scaled by $(1-q)$ and $r$, we can write $$|\triangle AEF| = (1-q) r\;|\triangle ABC| \tag{2}$$ Likewise, $$|\triangle BFD| = (1-r)p\;|\triangle ABC| \qquad\qquad |\triangle CDE| = (1-p)q\;|\triangle ABC| \tag{3}$$ so that \begin{align} |\triangle DEF| &= |\triangle ABC| - |\triangle AEF| - |\triangle BFD| - |\triangle CDE| \\[4pt] &=|\triangle ABC|\;\left(1-(1-q)r-(1-r)p-(1-p)q\right) \\[4pt] &=|\triangle ABC|\;\left( 1 - p - q - r + p q + p r + q r \right) \\[4pt] &=|\triangle ABC|\;\left(\; (1-p)(1-q)(1-r) + p q r \;\right) \tag{4} \end{align}

Observe that $(4)$ is obviously unchanged under the substitutions $$p \leftrightarrow 1-p \qquad q \leftrightarrow 1-q \qquad r \leftrightarrow 1-r$$

This implies that,

If $D$, $E$, $F$, $D^\prime$, $E^\prime$, $F^\prime$ are such that $$\overline{BD} \cong \overline{D^\prime C} \qquad \overline{CE} \cong \overline{E^\prime A} \qquad \overline{AF} \cong \overline{F^\prime B} \tag{\star}$$ then $$|\triangle DEF| = |\triangle D^\prime E^\prime F^\prime| \tag{\star\star}$$

(Note: To avoid marking overlapping segments, the diagram depicts $$\overline{BD^\prime} \cong \overline{DC} \qquad \overline{CE^\prime} \cong \overline{EA} \qquad \overline{AF} \cong \overline{F^\prime B}$$ but clearly these conditions are equivalent to $(\star)$.)

For the problem at hand, one needs only show that the points of contact of $\triangle ABC$'s edges with its incircles and excircles make a collection of points $D$, $E$, $F$, $D^\prime$, $E^\prime$, $F^\prime$ satisfying $(\star)$. Well, this certainly looks true:

Proof is not too difficult. See, for instance, the first part of this answer. $\square$