Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral with $A B \neq B C$. Denote by $\omega_{1}$ and $\omega_{2}$ the incircles of triangles $A B C$ and $A D C$. Suppose that there exists a circle $\omega$ inscribed in angle $A B C$, tangent to the extensions of line segments $A D$ and $C D$. Prove that the common external tangents of $\omega_{1}$ and $\omega_{2}$ intersect on $\omega$.

Solution:
The proof below is based on two known facts.

Lemma 1. Given a convex quadrilateral $A B C D$, suppose that there exists a circle which is inscribed in angle $A B C$ and tangent to the extensions of line segments $A D$ and $C D$. Then $A B+A D=C B+C D$.

Proof. The circle in question is tangent to each of the lines $A B, B C, C D, D A$, and the respective points of tangency $K, L, M, N$ are located as with circle $\omega$ in the figure. Then

$$
A B+A D=(B K-A K)+(A N-D N), \quad C B+C D=(B L-C L)+(C M-D M) .
$$

Also $B K=B L, D N=D M, A K=A N, C L=C M$ by equalities of tangents. It follows that $A B+A D=C B+C D$.

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For brevity, in the sequel we write "excircle $A C$ " for the excircle of a triangle with side $A C$ which is tangent to line segment $A C$ and the extensions of the other two sides.

Lemma 2. The incircle of triangle $A B C$ is tangent to its side $A C$ at $P$. Let $P P^{\prime}$ be the diameter of the incircle through $P$, and let line $B P^{\prime}$ intersect $A C$ at $Q$. Then $Q$ is the point of tangency of side $A C$ and excircle $A C$.

Proof. Let the tangent at $P^{\prime}$ to the incircle $\omega_{1}$ meet $B A$ and $B C$ at $A^{\prime}$ and $C^{\prime}$. Now $\omega_{1}$ is the excircle $A^{\prime} C^{\prime}$ of triangle $A^{\prime} B C^{\prime}$, and it touches side $A^{\prime} C^{\prime}$ at $P^{\prime}$. Since $A^{\prime} C^{\prime} \| A C$, the homothety with centre $B$ and ratio $B Q / B P^{\prime}$ takes $\omega_{1}$ to the excircle $A C$ of triangle $A B C$. Because this homothety takes $P^{\prime}$ to $Q$, the lemma follows.



Recall also that if the incircle of a triangle touches its side $A C$ at $P$, then the tangency point $Q$ of the same side and excircle $A C$ is the unique point on line segment $A C$ such that $A P=C Q$.

We pass on to the main proof. Let $\omega_{1}$ and $\omega_{2}$ touch $A C$ at $P$ and $Q$, respectively; then $A P=(A C+A B-B C) / 2, C Q=(C A+C D-A D) / 2$. Since $A B-B C=C D-A D$ by Lemma 1, we obtain $A P=C Q$. It follows that in triangle $A B C$ side $A C$ and excircle $A C$ are tangent at $Q$. Likewise, in triangle $A D C$ side $A C$ and excircle $A C$ are tangent at $P$. Note that $P \neq Q$ as $A B \neq B C$.

Let $P P^{\prime}$ and $Q Q^{\prime}$ be the diameters perpendicular to $A C$ of $\omega_{1}$ and $\omega_{2}$, respectively. Then Lemma 2 shows that points $B, P^{\prime}$ and $Q$ are collinear, and so are points $D, Q^{\prime}$ and $P$.

Consider the diameter of $\omega$ perpendicular to $A C$ and denote by $T$ its endpoint that is closer to $A C$. The homothety with centre $B$ and ratio $B T / B P^{\prime}$ takes $\omega_{1}$ to $\omega$. Hence $B, P^{\prime}$ and $T$ are collinear. Similarly, $D, Q^{\prime}$ and $T$ are collinear since the homothety with centre $D$ and ratio $-D T / D Q^{\prime}$ takes $\omega_{2}$ to $\omega$.

We infer that points $T, P^{\prime}$ and $Q$ are collinear, as well as $T, Q^{\prime}$ and $P$. Since $P P^{\prime} \| Q Q^{\prime}$, line segments $P P^{\prime}$ and $Q Q^{\prime}$ are then homothetic with centre $T$. The same holds true for circles $\omega_{1}$ and $\omega_{2}$ because they have $P P^{\prime}$ and $Q Q^{\prime}$ as diameters. Moreover, it is immediate that $T$ lies on the same side of line $P P^{\prime}$ as $Q$ and $Q^{\prime}$, hence the ratio of homothety is positive. In particular $\omega_{1}$ and $\omega_{2}$ are not congruent.

In summary, $T$ is the centre of a homothety with positive ratio that takes circle $\omega_{1}$ to circle $\omega_{2}$. This completes the solution, since the only point with the mentioned property is the intersection of the the common external tangents of $\omega_{1}$ and $\omega_{2}$.